Chapter 4math.nelson.com/calculusandvectors/Resources/Chapter4-6.pdf · c. i. ii. iii. d. i. ii....

Review Exercise, pp. 156–159

1. ,

2.

3. ,

4. ,

5. The upward velocity is positive forzero for and

negative for

6. a. min: ,max: 0

b. min: ,max: 16

c. min: 12,max: 20

7. a. 62 mb. Yes, 2 m beyond the stop sign c. Stop signs are located two or more

metres from an intersection. Sincethe car only went 2 m beyond thestop sign, it is unlikely the carwould hit another vehicle travellingperpendicular.

8. min is 2, max is 9. 250

10. a. i. $2200ii. $5.50iii. $3.00; $3.00

b. i. $24 640ii. $61.60

iii. $43.20; $43.21c. i. $5020

ii. $12.55iii. $0.03; $0.03

d. i. $2705ii. $6.76

iii. $4.99; $4.9911. 200012. a. moving away from its starting point

2 � 3�3

�65

�52Time

Velo

city

(m/

s)

08

30

45

15

–15

–30

–45

642 10

v(t)

t

t 7 4.5 s.t � 4.5 s,0 � t 6 4.5 s,

a1t 2 � 10t�3

v1t 2 � 1 � 5t�2

a � 2 � 12t � 3 2�32

v � 2t � 12t � 3 2�12

d2y

dx2� 72x7 � 42x

f – 1x 2 � 12x2 � 20x�6

f ¿ 1x 2 � 4x3 � 4x�5

A n s w e r s 641NEL

b. moving away from the origin andtowards its starting position

13. a.

b. Yes, since for all ,the particle is accelerating

14. 27.14 cm by 27.14 cm for the base andheight 13.57 cm

15. length 190 m, width approximately 63 m16. 31.6 m by 11.6 m by 4.2 m17. radius 4.3 cm, height 8.6 cm18. Run the pipe 7.2 km along the river shore

and then cross diagonally to the refinery.19. 10:35 p.m.20. $204 or $206 21. The pipeline meets the shore at a point C,

5.7 km from point A, directly across from P.22. 11.35 cm by 17.02 cm23. The two identical brick sides should have

length 25 m; the fenced side and thecorresponding brick side should havelength 40 m.

24. 2:23 p.m.25. 3.2 km from point C26. a. absolute maximum:

absolute minimum:b. absolute maximum:

absolute minimum:c. absolute maximum:

absolute minimum:d. absolute maximum:

absolute minimum:27. a. 62.9 m c.

b. 4.7 s

28. a. d.

b. e.

c. f.

29. a. position: 1, velocity: ,

acceleration: speed:

b. position: , velocity: ,

acceleration: , speed:

30. a.

b.c.d. undefined e.

Chapter 3 Test, p. 160

1. a.b.c.d. f – 1x 2 � 9614x � 8 2y– � 60x�5 � 60x

f – 1x 2 � �180x3 � 24xy– � 14

0.141 m>s2

2.36 m>s1.931 m>sa1t 2 �2

9 1t2 � t 2�4

3 12t2 � 2t � 1 2v1t 2 �

2

3 1t2 � t 2�1

3 12t � 1 2 ,49

1027

49

83

16�Q 1

18R,16

f – 18 2 � �1

72f – 10 2 � 192

f – 14 2 � � 1

108f – 1�1 2 � 26

f – 11 2 � �5

16f – 12 2 � 60

3.6 m>s2f 1�2 2 � �56f 14 2 � 2752,f 1�5 2 � �63f 15 2 � 67,f 1�3 2 � �18f 13 2 � 36,f 11 2 � 5f 17 2 � 41,

t 7 0a 7 0

t �2

3

2. a.

b.

3. a.

b. mc.d. between and e. 2 m s

4. a. min: max: 67 b. min: 6, max: 10

5. a. 2.1 sb.

6. 250 m by 166.7 m7. 162 mm by 324 mm by 190 mm8.

Chapter 4

Review of Prerequisite Skills, pp.162–163

1. a. or

b. or

c.

d. or or

2. a.

b.c.d. or

3. a.

b. y

x

4

6

2

8

0

–4

–2

–8

–6

–12

–10

–4–6 –2–8 4 62 8

y

x

2

3

1

0

–2

–3

–1–2 –1–3 2 31

x 7 1x 6 �4�1 6 t 6 3x � 2

x 6 �7

3

y � �2y � �3y � 1

x � �5

2

x � �2x � 7

y � 1y � �3

2

$850>month

about 22.9 m

�63,> t � 1.5 st � 0 s

1 m>s�0.25a1t 2 � 2v1t 2 � 2t � 3,a12 2 � �24v12 2 � 6,a13 2 � �44v13 2 � �57,

c.

d.

4. a. 0b. 7c. 27d. 3

5. a.

b.

c.

d.

6. a.

b.

7. ,

8. a. If where n is a realnumber, then

b. If where k is a constant,then

c. If then

d. If then

e. If f and g are functions that havederivatives, then the compositefunction has aderivative given by

f. If u is a function of x, and n is apositive integer, then

9. a. As .

b. AsAs

c. AsAs

d. AsAs

10. a.

b.

c. no vertical asymptote

d.

11. a.b.

c.

d.12. a. i. no x-intercept; (0, 5)

ii. (0, 0); (0, 0)

iii.

iv. ; no y-intercept

b. i. Domain: ,Range:

ii. Domain: ,Range:

iii. Domain: ,

Range:

iv. Domain: ,Range:

Section 4.1, pp.169–1711. a. (0, 1),

b. (0, 2)

c.

d. ,

2. A function is increasing whenand is decreasing when

3. a. i.ii.

iii.b. i.

ii.iii. (1, 4)

c. i.ii.

iii. noned. i.

ii.iii. (2, 3)

4. a. increasing: ;decreasing:

b. increasing: ;decreasing:

c. increasing: ;decreasing:

d. increasing: ;decreasing:

e. increasing: ;decreasing:

f. increasing: ;decreasing:

5. increasing: ;decreasing:

6.

7.8.

9. a. i.ii.

iii.

b. i.ii.

iii.y

x

2

3

1

0

–2

–3

–1–2 –1–3 2 31

x � �1, x � 1�1 6 x 6 1x 6 �1, x 7 1

y

x

2

3

1

0

–2

–3

–1–1 2 3 4 51

x � 4x 7 4x 6 4

c � �9b � �9,a � 3,

0

2

1

–2

–1

3

4

5

y

x

21 4 53–2 –1(–1, 0)

(2, 5)

x 6 �3, �2 6 x 6 1�3 6 x 6 �2, x 7 1

x 6 0x 7 0

0 6 x 6 1x 6 �2,x 7 1�2 6 x 6 0,

x 6 �1, x 7 3�1 6 x 6 3

0 6 x 6 1�1 6 x 6 0,x 6 �1, x 7 10 6 x 6 4x 6 0, x 7 4�2 6 x 6 0x 6 �2, x 7 0

2 6 x 6 3x 6 �1,3 6 x�1 6 x 6 2,

2 6 x�2 6 x 6 2,x 6 �21�1, 2 2 , x 7 1x 6 �1,�1 6 x 6 1

12, �1 21�1, 4 2 ,�1 6 x 6 2x 7 2x 6 �1,

f ¿ 1x 2 6 0.f ¿ 1x 2 7 0

a�1, �5

2ba1,

5

2b

1�2, �125 212.25, �48.2 2 ,a1

2, 0b ,

1�4, 33 25y�R 0y � 265x�R 0x � 06e y�R 0y �

1

2f

e x�R 0x �1

2f

5y�R 0y � 465x�R 0x � 265y�R 0y � 065x�R 0x � �16Q25, 0Ra 5

3, 0b ; a0,

5

3b

y � 2

y �12

y � 4y � 0

x � �311x � 3 2 2;11x � 4 2 2 � 1;

x � 31�x � 3;

x � 012x;

xSq, f 1x 2 Sq.xS�q, f 1x 2 S�q.xSq, f 1x 2 S�q.xS�q, f 1x 2 S�q.

xSq, f 1x 2 S �q.xS�q, f 1x 2 Sq.

xS ;q, f 1x 2 Sqd

dx1un 2 � nun�1du

dx.

h¿ 1x 2 � f ¿ 1g1x 2 2 # g¿ 1x 2 .h1x 2 � f 1g1x 2 2

g1x 2 � 0.

�f ¿ 1x 2g1x 2 � f 1x 2g¿ 1x 23g1x 2 42 ,h¿ 1x 2

h1x 2 �f 1x 2g1x 2 ,

� f 1x 2g¿ 1x 2k¿ 1x 2 � f ¿ 1x 2g1x 2k1x 2 � f 1x 2g1x 2 ,f ¿ 1x 2 � 0.f 1x 2 � k,

f ¿ 1x 2 � nxn�1.f 1x 2 � xn,

1�1, 4.5 2a 2

3, 2.19bx � 7 �

2

x � 1

x � 8 �28

x � 3

t � 81t � 4 2 32213x2 � 6x 2 16x � 6 2�

x2 � 2x � 31x2 � 3 22x3 � 6x � x�2

y

x

6

4

2

0

–4

–6

–2–4 –2–6 4 62

y

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

A n s w e r s642 NEL

0

2

–2

–4

–6

–8

4

6

8

–2 2 4 6 8–4–6–8

(1, 2)

(–5, 6)

y

x

c. i.ii.

iii.

d. i.ii.

iii.

10.

Let then

If therefore the

function is decreasing.If therefore thefunction is increasing.

11. ,increasing: ,decreasing: ,local minimum:

12.

13. Let and Let and be any two values in theinterval so that Since both functions areincreasing:

(1)(2)

results in

The function yu or is strictlyincreasing.

14. strictly decreasing

Section 4.2, pp. 178–180

1. Determining the points on the graph ofthe function for which the derivative ofthe function at the x-coordinate is 0 oris undefined.

2. a. Take the derivative of the function.Set the derivative equal to 0. Solvefor x. Evaluate the original functionfor the values of x. The (x, y) pairsare the critical points.

b.

3. a. local minima: ,local maximum:

b. local maximum: ,local minimum: (3, 0.3)

c. local minimum: ,local maximum: (0, 1)

4. a.

b.

c. (0, 1)

5. a. local minimum: (0, 3),local maximum: (2, 27),Tangent is parallel to the horizontalaxis for both.

b. (0, 0) neither maximum norminimum,Tangent is parallel to the horizontalaxis.

c. (5, 0); neither maximum norminimum,Tangent is not parallel to thehorizontal axis.

d. local minimum: ,Tangent is parallel to the horizontalaxis.

and (1, 0) are neithermaxima or minima.Tangent is not parallel to thehorizontal axis for either.

6. a.

1�1, 0 210, �1 2

0

10

–10

–20

20

30

t

2 4 6–2

h(t)

1�3.1, 0 2 ,

y

x

0.5

0

–0.5

–2–4 2 4

10, 0 2

y

x

10

20

0

–10

–20

–2–4 2 4

10, 0 2 , 12�2, 0 2 , 1�2�2, 0 21�2, 5 21�3, �0.3 210, 0 2 12, �16 21�2, �16 2 ,

y

x

20

0

–20

–40

–4 4 8

14, �32 210, 0 2 ,

f(x)

g(x)

a bx1 x1

y

x

f 1x 2 # g1x 2f 1x2 2 # g1x2 2 7 f 1x1 2g1x1 211 2 � 12 2

yu � f 1x 2 # g1x 2g1x2 2 7 g1x1 2f 1x2 2 7 f 1x1 2x1 6 x2,

x1 6 x2.a � x � bx2x1

u � g1x 2 .y � f 1x 2

y

x

4

8

0

–4

–2 2 4

12, �44 2x 6 2x 7 2

f ¿ 1x 2 � 0 for x � 2

f ¿ 1x 2 7 0,x 7 �b2a ,

f ¿ 1x 2 6 0,x 6 �b2a ,

x ��b2a .f ¿ 1x 2 � 0,

f ¿ 1x 2 � 2ax � bf 1x 2 � ax2 � bx � c

0

2

1

3

4

5

y

x

21 4 53

x � 2x 6 2x 7 2

y

x

2

3

4

5

1

0–1

–2 –1–3 2 31

x � �2, x � 3x 6 �2, x 7 3�2 6 x 6 3


b.

c.

d.

7. a. local maximum

b. local maximum,local minimum

c. local maximum,local minimum

d. no critical points

e. local minimum

f. (0, 0) neither maximum norminimum,

local minimum

8. local minima at and local maximum at

9.

10.

11.minimum; the derivative is negative to the left and positive to the right

12. a.b.c.

13.14. a.

b.

c.

d.

15. a. , , ,b.c. local minimum: and

,local maximum:

16. a. local maximum: (0, 4)

x

4

2

0

–4

–2–4 –2 42

y

10, �9 213, �198 2 1�2, �73 213, �198 2 d ��9c � 0b � �36a � �4

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

f '(x)

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

f '(x)

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

f '(x)

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

f '(x)

d � 0c � 0,b � 3,a � �1,k 7 0k � 0k 6 0

q � 6p � �2,

c � 1b �22

3,a � �

11

9,

f(x)

x

2

3

4

5

6

1

0–1

–2

–2 –1–3–4 2 3 41

(3, 1)

(–1, 6)

x � �1x � 2;x � �6

y

x

4

8

0–2–4 2 4

11, �1 2

y

x

4

8

0–2–4 2 4

11, 1 2

y

x

4

8

0

–4

–8

–1–2 1 2

y

x

4

8

0

–4

–8

–2–4 2 4

1�1, �5 21�2, �4 2

y

x

10

20

0

–10

–20

–4–8 4 8

13, �16 21�3, 20 2

y

x

10

20

0

–10

–20

–4–8 4 8

12, 21 2


b. local minimum: ,local maximum:

17.

Since has a local maximum atthen for and

for Since has alocal minimum at then

for and for

If and thenIf and thenSince for , and for

Therefore, hasa local maximum at

Section 4.3, pp. 193–195

1. a. vertical asymptotes at andhorizontal asymptote at

b. vertical asymptote at horizontal asymptote at

2.

Conditions for a vertical asymptote:must have at least one

solution s, and

Conditions for a horizontal asymptote:where or

, where

Condition for an oblique asymptote:The highest power of must be onemore than the highest power of

3. a. 2b. 5

c.

d. ,4. a. large and positive to left of

asymptote, large and negative toright of asymptote

b. large and negative to left ofasymptote, large and positive toright of asymptote

c. large and positive to left ofasymptote, large and positive toright of asymptote

d. hole at no vertical asymptotee. large and positive to left of

asymptote, large and negative toright of asymptote

large and negative to left ofasymptote, large and positive toright of asymptote

f. large and positive to left ofasymptote, large and negative toright of asymptote


5. a. large negative: approachesfrom above, large positive:approaches from below

b. large negative: approachesfrom below, large positive:approaches from above

c. large negative: approachesfrom above, large positive:approaches from above

d. no horizontal asymptotes6. a.

b.

c.

d.

7. a.b.c.d.

8. a. large negative: approaches frombelow, large positive: approachesfrom above

b. large negative: approaches fromabove, large positive: approachesfrom below

9. a. large and positive to leftof asymptote, large and negative toright of asymptote

b. large and positive to left ofasymptote, large and positive toright of asymptote

c. large and negative to leftof asymptote, large and positive toright of asymptote

d. large and negative to left ofasymptote, large and positive toright of asymptote

10. a.f(x)

x

2

4

0

–2

–4

–2–4 2 4

x � 2;y � 1

x � �2;y � 1

x � 1;y � 3

x � �5;

y � x � 3y � x � 2y � x � 3y � 3x � 7

y

x

4

2

6

8

0

–4

–2

–6

–8

–4 –2–6–8 42 6 8

0

2

–2

–4

–6

–8

4

6

8

–2 2 4 6–4–6–8

y

x

y

8

10

6

4

2

0–2

–2 2–4–6–8

x

y

4

6

2

0

–4

–2

–6

–2–4–6–8–10

x

y � 3;

y � 0;

y � 1;

x � 1;

x � �1;

x � 1;

x � �3;x � 3,

x � 3;

x � 2;

x � �5;�qq

�5

2

k1x 2 .g1x 2k�R.lim

xS�qf 1x 2 � k

k�R,limxSq

f 1x 2 � k,

limxSq

f 1x 2 � q.h1x 2 � 0

f 1x 2 �g1x 2h1x 2

y � 0x � 0;

y � 1x � 2;x � �2

x � c.h1x 2h¿ 1x 2 6 0.x 7 c,

h¿ 1x 2 7 0x 6 ch¿ 1x 2 6 0.

g¿ 1x 2 7 0,f ¿ 1x 2 6 0x 7 c,h¿ 1x 2 7 0.

g¿ 1x 2 6 0,f ¿ 1x 2 7 0x 6 c,

h¿ 1x 2 �f ¿ 1x 2g1x 2 � g¿ 1x 2 f 1x 23g1x 2 42

h1x 2 �f 1x 2g1x 2

x 7 c.g¿ 1x 2 7 0x 6 cg¿ 1x 2 6 0

x � c,g1x 2x 7 c.f ¿ 1x 2 6 0

x 6 cf ¿ 1x 2 7 0x � c,f 1x 2

h1x 2 �f 1x 2g1x 2

x

40

20

0

–40

–20–4 –2 42

y

1�1.41, 39.6 211.41, �39.6 2


b.

c.

d.

e.

f.

11. a.

b.

12. a.

b.

13. a.

b.

14. a. and : and exist.

b. : the highest degree of x in thenumerator is exactly one degreehigher than the highest degree of xin the denominator.

c. : the denominator is defined forall .

has vertical

Asymptotes at and .As .As .As .As .

has a horizontal asymptote at

has a vertical asymptoteatAs .As .

is an oblique asymptote for.

has vertical

asymptotes at and .As .As .As .As .

has a horizontal asymptote at

15.

16. a.

b.

17.

Mid-Chapter Review,pp.196–197

1. a. decreasing: ,increasing:

b. decreasing: (0, 2),increasing:

c. increasing: ,d. decreasing: ,

increasing: 10, q 21�q, 0 2 1�3, q 21�q, �3 21�q, 0 2 , 12, q 212, q 21�q, 2 2

f (x)

x1

4

3

2

0–1

–2

–3

–4

–2–4–6–8–10 2 4 6 8 10

� limxSq

�2� �2

� limxSq

�2x

x � 1

� limxSq

x2 � 1 � x2 � 2x � 1

x � 1

limxSqc x2 � 1

x � 1�

x2 � 2x � 1

x � 1d

� q

� limxSq1x � 1 2� lim

xSq

1x � 1 2 1x � 1 21x � 1 2limxSq

x2 � 2x � 1

x � 1

� q

limxSq

x2 � 1

x � 1� lim

xSq

x �1x

1 �1x

b �3

5a �

9

5,

y � 1.r 1x 2xS 4�, r1x 2 SqxS 4�, r1x 2 S�q

xS�4�, r1x 2 S�qxS�4�, r1x 2 Sq x � 4x � �4

r 1x 2 �1x � 3 2 1x � 2 21x � 4 2 1x � 4 2

h1x 2y � xg1x 2 S�qxS 3�,g1x 2 SqxS 3�,

x � 3.g1x 2

y � 0.f 1x 2 f 1x 2 S�qxS 7�,

f 1x 2 SqxS 7�,f 1x 2 SqxS�2�,f 1x 2 S�qxS�2�,

x � �2x � 7

f 1x 2 ��x � 31x � 7 2 1x � 2 2

x�Rh 1x 2

h 1x 2limxSq

r 1x 2limxSq

f 1x 2r 1x 2f 1x 2

x

2

4

0

–2

–4

–2 2 4 6

y

0

2

1

–2

–1

y

x

42–4 –2

0

4

2

–2

6

8

y

x

42–4 –2

y = f ''(x)

0

4

2

–4

–6

–2

6

y

x

42–4 –2

y = f ''(x)

x � �d

c

y �a

c

s(t)

t2

4

6

8

10

0–2

–2–4–6–8 2 4 6

g(x)

x

8

16

0

–8

–16

–8–16 8 16

t

2

4

0

–2

–4

–2–4 2 4

s(t)

y

x

2

4

0

–2

–4 –2–6–8 4 6 82

4 60

20

10

30

–10

–20

h(t)

t

2–2


2. increasing: and ,decreasing:

3.

4. a. d.b. e. 0c. 0, f.

5. a. local maximum,local minimum

b. local maximum,

local minimum6.

7. local minimum,increasing: ,decreasing:

8. a. large and positive to left ofasymptote, large and negative toright of asymptote

b. large and negative to leftof asymptote, large and positive toright of asymptote

large and positive to left ofasymptote, large and negative toright of asymptote

c. large and negative to leftof asymptote, large and positive toright of asymptote

d. large and positive to left ofasymptote, large and negative toright of asymptote

large and positive to left ofasymptote, large and negative toright of asymptote

9. a. large negative: approachesfrom above, large positive:approaches from below

b. large negative: approachesfrom below, large positive:approaches from above

10. a. large and positive to left ofasymptote, large and positive toright of asymptote

b. no discontinuitiesc. ; large and negative

to left of asymptote, large andpositive to right of asymptote

large and negativeto left of asymptote, large andpositive to right of asymptote

11. a. is increasing.b. is decreasing.

12. a. increasing: ,decreasing:

b. The velocity is always decreasing.13. increasing: ,

decreasing:14.

15. a. i.

ii. increasing: ,

decreasing:

iii. local minimum at

iv.

b. i.ii. increasing: ,

decreasing:iii. local minimum at

local maximum at iv.

c. i.ii. increasing: ;

decreasing:

iii. local maximum at local minimum at

iv.

d. i.ii. increasing: ;

decreasing:iii. local maximum at

local minimum at iv.

16. a. ; large and positive to left of asymptote, large and negative to

right of asymptote; ; As

, the function approaches

the horizontal asymptote from above.

As , the function approaches

the horizontal asymptote from

below.b. large and positive to left

of asymptote, large and positive toright of asymptote; ; As

, the function approachesthe horizontal asymptote from above.As , the function approachesthe horizontal asymptote frombelow.

c. large and positive to left of asymptote, large and negative toright of asymptote; ; As

, the function approachesthe horizontal asymptote from above.As , the function approachesthe horizontal asymptote frombelow.

d. large and negative to leftof asymptote, large and positive toright of asymptote

x � �4;

xSq

xS�qy � �1

x � �3;

xSq

xS�qy � 1

x � �2;

xSq

xS�qy �

12

x � �12

y

x

4

2

6

8

0

–4

–2

–6

–8

–1–2 1 2

x � 1x � �1,

�1 6 x 6 1x 7 1x 6 �1,

x � 1x � �1,

y

x

2

1

3

0

–1

–1–2 1 2

x � 1, �1x � 0,

0 6 x 6 1x 6 �1,x 7 1�1 6 x 6 0,

x � 1x � 0,x � �1,

y

x

25

20

15

10

5

30

35

0–5

–2 42 6

x � 3x � 0,x 7 3x 6 0,

0 6 x 6 3x � 3x � 0,

y

x

10

5

0

–20

–15

–10

–25

–30

–5–4 –2 82 4 6 10 12

x �72

x 6 72

x 7 72

x �7

2

y

x

5

0

–10

–15

–5

–2–4 2 4

t 6 2.5198t 7 2.5198

t 7 0.970 6 t 6 0.97

f 1x 2f 1x 2

x � 6 � 2�6;

x � 6 � 2�6

x � 5;

y � 1;

y � 3;

x � 5;

x � �23;

x � �3;

x � 3;

x � �3;

x � �2;x 6 2x 7 2

x � 2

;2x � 2

x � �23

x � 2x � 1

;�2;�2;3

;2;1,�4

0

2

–2

–4

–6

–8

4

6

8

–2 2 4 6 8–4–6–8

(3, 5)

(–2, 0)

y

x

�1 6 x 6 2x 7 2x 6 �1


17. a. e.

b. f. 1

c. g. 1d. 0 h.

Section 4.4, pp.205–206

1. a. A: negative, B: negative, C: positive,D: positive

b. A: negative, B: negative, C: positive,D: negative

2. a. local minimum: ,local maximum:

b. local maximum:

c. local maximum: ,local minimum: (1, 2)

d. (3, 8) is neither a local maximum or minimum.

3. a.

b.

c. no points of inflectiond. (3, 8)

4. a. aboveb. 4; above

c. below

d. below

5. a. i. concave up on ,concave down on

ii.

iii.

b. i. concave up on or ,concave down on

ii. andiii.

6. For any function find thecritical points, i.e., the values of x suchthat or does not exist.Evaluate for each critical value.If the value of the second derivative at

a critical point is positive, the point is alocal minimum. If the value of thesecond derivative at a critical point isnegative, the point is a local maximum.

7. Use the first derivative test or thesecond derivative test to determine thetype of critical points that may bepresent.

8. a. i. (0, 0)ii.

b. i. ,

ii.

9.

10. , ,

11.

12.

For possible points of inflection, wesolve f – 1x 2 � 0:

f – 1x 2 � 12ax2 � 6bxf ¿ 1x 2 � 4ax3 � 3bx2f 1x 2 � ax4 � bx3

27

64

y

x

4

2

6

8

10

0–2

–2–4 42 6 8

(2, 11)

(1, 5)

c � �1b � 9a � �3

y

x2

4

0–2

–4

–2–4 2 4 6 8

y

x

2

1

3

4

0

–2

–1

–3

–4

–2 –1–3–4 21 3 4

Q 3�2,

8�29 RQ� 3

�2, �8�2

9 R

y

x

10

5

15

20

0

–10

–5

–15

–20

–25

–2–4 2

1�2, �16 2 ,

f – 1x 2 f ¿ 1x 2f ¿ 1x 2 � 0

y � f 1x 2 ,

y

x

2

4

0

–2

–2 2 4

y = f(x)

x � 2x � 00 6 x 6 2

x 7 2x 6 0

0

2

1

–1

3

4

y

x

1 2–1–2

y = f(x)

x � 1x 7 1

x 6 1

�227;

�9

100�10;

24;

Q4, 2564RQ�4, 25

64R12, �36 2

1�1, �2 2Q0, 2548R

1�1, 18 215, �90 2

q�3

1

6

q�2

3

or

The graph of is a parabola

with x-intercepts 0 and

We know the values of have opposite signs when passing through a root. Thus, at and at the concavity changes as the graph goes through these points.Thus, has points of inflection at

and

To find the x-intercepts, we solve

or

The point midway between the

x-intercepts has x-coordinate

The points of inflection are (0, 0) and

13. a.

b. Answers may vary. For example,there is a section of the graph thatlies between the two sections of thegraph that approaches the asymptote.

14. If is odd but not 1, then there is aninflection point at . If is even,there is not an inflection point.

Section 4.5, pp. 212–213

1. A cubic polynomial that has a localminimum must also have a localmaximum. If the local minimum is tothe left of the local maximum, then

as andas If the local

minimum is to the right of the localmaximum, then as

and as 2. A polynomial of degree three has at

most two local extremes. A polynomialof degree four has at most three localextremes. Since each local maximumand minimum of a function corresponds

xS�q.f 1x 2 S�qxS�qf 1x 2 S�q

xS�q.f 1x 2 S�qxS�qf 1x 2 S�q

nx � cn

y

x

8

4

0–4

–8

–12

–16

–2–4–6 2 4 6

Q� b2a, �

b4

16a3R.�

b2a.

x � �bax � 0

x31ax � b 2 � 0

f 1x 2 � 0

x � �b2a.x � 0

f 1x 2x � �

b2a,

x � 0

f – 1x 2�b2a.

y � f – 1x 2x � �b

2a.x � 0

6x12ax � b 2 � 012ax2 � 6bx � 0


to a zero of its derivative, the numberof zeros of the derivative is themaximum number of local extremevalues that the function can have. For apolynomial of degree n, the derivativehas degree so it has at most

zeros, and thus at most local extremes.

3. a. orb. no vertical asymptotesc.

4. a.

b.

c.

d.

e.

f.

g.

h.

i.

j.

5. a.

b.

c.

d.

e.

f.

y

x

4

2

6

8

0

–4

–2–2–6 –4 862 4 10

y

x2

4

6

8

0–2

–4

–6

–8

–4–6 –2–8 4 62 8

(3, 3)

(–1, –5)

y

x

1.0

2.0

0

–1.0

–2.0

–2–4–6–8 2 4 6 8

y

4

8

0

–8

–12

–2–4–6 2 4 6–4

x

y

x

4

2

6

8

0

–4

–2

–6

–8

–2–4 862 4 10 12

y

x

2

1

3

0–1

–2

–3

–8 –4–12 84 12

y

x

20

40

0

–20

–40

–60

–80

–2 –1–3 21 3 4

(2, 48)

(3, 45)

(–2, –80)

y

x

4

8

0

–4

–8

–2–4 2 4

y

x

15

30

45

60

0–2 2 4 6 8 10

(3, 57)

( , 30)23

y

x

15

30

0

–15

–30

–2 2 4 6 8 10

(5, 5)

(3, 21)

(1, 37)

x � 3

x � �1x � �3

n � 1n � 1n � 1,


g.

h.

i.

j.

6.

7. a. Answers may vary. For example:

b. Answers may vary. For example:

8.

9.

10. a. is a horizontal asymptote tothe right-hand branch of the graph.

is a horizontal asymptoteto the left-hand branch of the graph.

b. and are horizontal

asymptotes.11.

For possible points of inflection, we

solve

The sign of changes as x goes from

values less than to values greater

than Thus, there is a point of

inflection at

At


1. a. i.ii.

iii. (1, 20)b. i.

ii.iii.

2. No, a counter example is sufficient tojustify the conclusion. The function

is always increasing, yet thegraph is concave down for andconcave up for

3. a. (0, 20), local minimum; tangent ishorizontal(3, 47), local maximum; tangent ishorizontal

b. (0, 6), local maximum; tangent ishorizontal(3, 33), neither local maximum norminimum; tangent is horizontal

c. , local minimum;

, local maximum; tangents at

both points are paralleld. (1, 0), neither local maximum nor

minimum; tangent is not horizontal4. a.

b.c.d.

5. a. ; large and negative to left ofasymptote, large and positive toright of asymptote

b. ; large and positive to left of asymptote, large and negative toright of asymptote

c. hole at d. large and positive to left

of asymptote, large and negative toright of asymptote


6. (0, 5); Since the derivative is 0 at ,the tangent line is parallel to the x-axisat that point. Because the derivative isalways positive, the function is alwaysincreasing and, therefore, must cross thetangent line instead of just touching it.

x � 0

x � 5;

x � �4;x � �3

x � �5

x � 3c 6 x 6 dx 6 a, d 6 x 6 eb 6 x 6 ca 6 x 6 b, x 7 e

Q7, 114R

Q�1, �12R

x 7 0.x 6 0

f 1x 2 � x3

16.5, �1 211, �1 2 , 3 6 x 6 6.51 6 x 6 3,x 7 6.5

�3 6 x 6 1,x 6 �3,

x 7 1x 6 1

2b a�b

3ab � c � c�

b2

3a

dy

dx� 3a a�b

3ab 2

�x �b

3a,

x ��b3a .

�b3a .

�b3a

d2ydx2

x � �b

3a

d2ydx2 � 0:

d2y

dx2� 6ax � 2b � 6a a x �

b

3ab

dy

dx� 3ax2 � 2bx � c

y � ax3 � bx2 � cx � d

y � �32y �

32

y � �1

y � 1

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

y

(–1, –1.6)

y

x

2

1

3

0

–2

–1

–3

–4 –2 42 6 8

0

6

8

4

2

2

4

6

8

y

x

4 62 8–4–6–8 –2

0

6

8

10

4

2

2

4

6

8

10

y

x

4 62 8 10–4–6–8–10 –2

y

x

2

1

4

3

0

–2

–3

–1

–4

–2–3 –1–4 21 3 4

d � 0c � 3,b � 0,a � �1

4,


7.

8. a. i. concave up: ;concave down:

ii. points of inflection: ,

iii.

b. i. concave up: ,

concave down: ,

ii. points of inflection:, and

iii.

9. a.b.

10. a.

b.

c.

d.

e.

f.

11. a. ,b. There are three different graphs that

result for values of k chosen.

y

x

2

4

0

–2

–4

–2–4 2 4

k � 2

y

x

2

4

0

–2

–4

–2–4 2 4

k � 0

x � ; k�2 � k � 2

y

x

4

8

0

–4

–4 4 8

(1.6, 0.3)

(4.4, 5.8)

y

x

2

4

0

–2

–4

–2–4 2 4

y

x

20

40

–20

–40

–2 4 620

y

x

2

4

0

–2

–4

–2–4 2 4

y

x

2

4

0

–2

–4

–2–4 2 4

y

x

4

8

0

–4

–8

–2–4 2 4

(2, –9)(–2, –9)

y

x

4

6

8

2

0

–4

–2–4 –2 4 62

b � 0a � 1,

y

x

5

10

–10

–5

–5–10 5 100

x � 5x � 1x � �4.5,

1 6 x 6 5x 6 �4.5

5 6 x�4.5 6 x 6 1

y

x

40

80

–40

–80

–4 8 1240

x � 3x � �1

x 6 �1, 3 6 x�1 6 x 6 3

y

x

4

6

8

10

2

0–2

–4

–6

–2 –1 2 3 41

(–2, 10)

(1, –6)

(3, 4)


For all other values of k, the graph willbe similar to the graph below.

12. a.b.

13. ;increasing: ;decreasing:

14. local maximum: ,local minimum: (1.107, 0.446),absolute maximum: (3, 24.5),absolute minimum:

15.

16. a. oblique asymptote,vertical asymptotes at

and horizontal asymptote at

vertical asymptotes at and horizontal asymptote at

vertical asymptote at b.

17. Domain: ;x-intercept: ;vertical asymptote: large andnegative to the left of the asymptote,large and positive to the right of theasymptote;no horizontal or oblique asymptote;increasing: ;decreasing: ;concave up: ;concave down: ;local minimum at (1.59, 7.56);point of inflection at

18. If is increasing, thenFrom the graph of for

If is decreasing, then. From the graph of for At a stationary

point, . From the graph, the zerofor occurs at . At ,

changes from negative to positve,so has a local minimum point there.If the graph of is concave up, thenis positive. From the slope of thegraph of is concave up for

If the graph of isconcave down, then is negative.From the slope of the graph of isconcave down for and

Graphs will vary slightly.

19. domain: ;x-intercept and y-intercept: ;vertical asymptote: large andpositive on either side of the asymptote;horizontal asymptote: ;increasing: ;decreasing: ;

concave down: ;concave up: ;local minimum at ;point of inflection:

20. a. Graph A is f, graph C is andgraph B is We know this becausewhen you take the derivative, thedegree of the denominator increasesby one. Graph A has a squared termin the denominator, graph C has acubic term in the denominator, andgraph B has a term to the power offour in the denominator.

b. Graph F is f, graph E is and graphD is We know this because thedegree of the denominator increasesby one degree when the derivative istaken.


1. a. or oror

b. or or

c. (0, 2),d.e.f. org.

2. a. or or

b. local maximum

local maximum

: local minimum3. y

x

4

6

2

0

–4

–6

–2–4 –2–6 4 62

(–1, 7)

(1, 4)

(3, 2)

13, �45 2Q12, 158 R:

Q�12, �

178 R:

x �1

2x � �

1

2x � 3

110, �3 21�8, 0 2 , 4 6 x 6 8�3 6 x 6 0f – 1x 2 7 0

x � 4x � �3,18, �4 21�6, �4 2 ,1�9, 2 2 ,4 6 x 6 8

�3 6 x 6 0�9 6 x 6 �6x 7 80 6 x 6 4

�6 6 x 6 �3x 6 �9

f –.f ¿

f –.f ¿,

0

2

–2

4

6

y

x

2 4–2–4

1�2, �1.11 21�1, �1.25 2�2 6 x 6 1, x 7 1x 6 �2

x 6 �1, x 7 1�1 6 x 6 1

y � 0

x � 1;10, 0 25x�R 0 x � 16

0

y

x

1 2–1

1

–1

2

–2

x 7 0.6.x 6 �0.6

ff ¿,f –

f�0.6 6 x 6 0.6.f

f ¿,f –f

ff ¿ 1x 2 x � 0x � 0f ¿ 1x 2x � 0

x 6 0.f ¿ 1x 2 6 0f ¿,f ¿ 1x 2 6 0

f 1x 2x 7 0.f ¿, f ¿1x 2 7 0

f ¿ 1x 2 7 0.f 1x 20

8

4

–8

–4

12

16

y

x

42 6–4 –2–6

1�2, 0 2�2 6 x 6 0

x 6 �2, x 7 0x 6 0, 0 6 x 6 1.59x 7 1.59

x � 0;�2

5x�R 0 x � 06

0

4

2

–4

–2

6

8

10

y

x

2 4–2–4

x � 2s1x 2 :y � 1x � 1;

x � �1r1x 2 :y � 0x � 3;

x � �1q1x 2 : y � 0.75xp1x 2 :

y

x

100

150

200

50

–100

–50

–150

–200

–4–8 4 80

1�4, �7 21�2.107, 17.054 20 6 x 6 2x 6 �2,

x 7 2�2 6 x 6 0,x � 2x � 0,x � �2,

y � 4x � 11y � x � 3

y

x

2

4

0

–2

–4

–2–4 2 4


4. hole at vertical asymptote at; large and negative to left of

asymptote, large and positive to rightof asymptote;

;Domain:

5.

6. There are discontinuities at and

is a vertical asymptote.

is a vertical asymptote.

The y-intercept is and x-interceptis .

is a local minimum and

is a local maximum.

is a horizontal asymptote.

7.

Chapter 5

Review of Prerequisite Skills,pp. 224–225

1. a. c.

b. 4 d.

2. a.

b.

c.

d.

e.

f.

3. a.

x-intercept:

b.

no x-intercept

4. a. b. c.

5. a. e.

b. f.

c. g.

d. h.

6. a. c. e.

b. d. f.

7. a. ,

b. ,

c. ,

d. ,is undefined

8. a. period: ,amplitude: 1

b. period: ,amplitude: 2

c. period: 2,amplitude: 3

d. period: ,

amplitude:

e. period:amplitude: 5

f. period:

amplitude:

9. a.

b.

10. a.LS

RS

Therefore,sec x csc x.

tan x � cot x �

�1

cos x sin x

�1

cos x �1

sin x

� sec x csc x�

11cos x 2 1sin x 2�

sin2 x � cos2 x1cos x 2 1sin x 2�

sin xcos x �

cos xsin x

� tan x � cot xtan x � cot x � sec x csc x

3p2

5p2

7p2

3

2

1

0– 1

–2

–3

2p 3ppp2

x

y

3p2

3

4

2

1

0 2ppp2

x

y

32

p,

2p,

27

p6

4p

p

tan ucos u � 0

cos u �1

�5

sin u � �2

�5

tan u ��5

2

sin u � ��5

3

tan u � �5

12

cos u � �12

13

�bab

a

bab

11p

6

p

6

5p

4�p

2

�2p

3

p

4

3p

22p

y

x

x

r

y

r

80–2

y

x

42 6–4 –2–8 –6

4

2

6

8

10

1�1, 0 2

2 30

3

2

1

–1

–2

–3

1–2–3 –1

x

y

loga T � b

log3 z � 8

log10 450 � w

logx 3 � 3

log41

16� �2

log5 625 � 4

9

4

1

9

1

9

y

x

4

6

8

2

0–2

–4

–6

–8

–2 –1–3–4 2 3 41

(0, 2)

(–2, 6)

c � 2b � 3,

y

x

4

6

8

2

0–2

–4

–6

–8

–4 –2–6–8–10 4 6 8 102

y � 0

1�1, �1 2Q�9, �19R

�5�

109

limxS3�

f 1x 2 � �q

limxS3�

f 1x 2 � q¶ x � 3

limxS3�

f 1x 2 � q

limxS3�

f 1x 2 � �q¶ x � �3

x � 3.x � �3

y

x20

–40

–60

–80

–100

–120

–140

–20–4 –2–6 4 6 8 1020

5x�R 0 x � �2, x � 36y � 1

x � 3x � �2;


b.

LS

RS

Therefore,

11. a.

b.

Section 5.1, pp. 232–234

1. You can only use the power rule whenthe term containing variables is in thebase of the exponential expression. Inthe case of the exponentcontains a variable.

2. a.b.c.d.

e.

f.

3. a.b.

c.

d.

e.

f.

4. a.

b.

c.

5. a.

b.

c. The answers agree very well; thecalculator does not show a slope ofexactly 0.5, due to internal rounding.

6.

7.

8. (0, 0) and

9. If then

and

10. a. ,

,

b.

11. a. ,

b. ,

c. ,

12. a. 31 000

b.

c. bacteria hd. 31 000 at time e. The number of bacteria is constantly

decreasing as time passes.

13. a.

b.

From a, which

gives Thus,

c. 40 m/sd.

14. a. i. eii. e

b. The limits have the same valuebecause as x , 0.

15. a.b.

16. or

17. a.

b.

c. Since

18. a. Four terms:Five terms:Six terms:Seven terms: 2.718 055

2.716 6662.708 3332.666 666

�11cosh x 22

�

1

4 14 2

1cosh x 22

�1e2x � 2 � e�2x 2 41cosh x 22

�

1

43 1e2x � 2 � e�2x 21cosh x 22

�

1

2 1ex � e�x 2 a 1

2b 1ex � e�x 2

1cosh x 22

�

1

2 1ex � e�x 2 a 1

2b 1ex � e�x 2

1cosh x 22

�

1sinh x 2 a d

dx cosh x b

1cosh x 22

d

dx1tanh x 2 �

a d

dx sinh x b 1cosh x 21cosh x 22

tanh x �sinh xcosh x,

� sinh x

d

dx1cosh x 2 �

1

21ex � e�x 2

� cosh x�

1

21ex � e�x 2

d

dx1sinh x 2 �

d

dxc 121ex � e�x 2 d

m � 2m � �3e21

S1xqS

about 12 s, about 327.3 m

a � 10 a1 �v

40b � 10 �

1

4v.

et4 � 1 �

v

40.

v � 40 Q1 � e� t4R,

a �dv

dt� 40 a 1

4e� t

4 b � 10e� t4

40 Q1 � e� t4R

t � 0>�17

�100

3e� t

30

d2y

dx2� ex12 � x 2

dy

dx� ex13 � x 2

d2y

dx2� 4xe2x � 4e2x

dy

dx� e2x12x � 1 2

d2y

dx2� �3ex

dy

dx� �3ex

dn y

dxn � 1�1 2n 13n 2 e�3x

d3y

dx3� �27e�3x

d2y

dx2� 9e�3x

dy

dx� �3e�3x

�1

25y

�1

25c 52a e

x5 � e�x

5 b dy– �

5

2a 1

25e

x5 �

1

25e�x

5 by¿ �

5

2a 1

5e

x5 �

1

5e�x

5 b ,

y �5

2 1ex

5 � e�x5 2 ,

a2,4

e2b

y �1

e

2 30

3

2

1

–1

–2

–3

1–2–3 –1

x

y

ex � y � 0

y �1

2x � 1

�2 � 3e

1

e

e3 � e�3

2e2t

11 � e2t 222tet2

� 3e�t

�xex � ex a 1

2�xb

�3x2e�x31x 2 � e�x3

x2

e3x13x � 1 26x2ex3

1

2�xe�x

1�6 � 2x 2e5 � 6x � x2

�3e�3x20e10t3e3t � 53e3x

y � ex,

5p

3x �p

3,

5p

6x �p

6,

sin x1 � sin2 x

� tan x sec x.

�sin xcos2 x

�sin xcos x �

1

cos x

� tan x sec x�

sin x

cos2 x

�sin x

1 � sin2 x

sin x1 � sin2 x

� 1tan x 2 1sec x 2


b. The expression for e in part a is a

special case of

in that it is the

case when Then is in fact

The value ofx is 1.

Section 5.2, p. 2401. a.

b.

c.

d.

e.

f.

2. a.

b.

c.

d.

3.

4.5.6. a. about years

b. about % year7. a. In 1978, the rate of increase of debt

payments was $904 670 000 annumcompared to $122 250 000 annumin 1968. The rate of increase for1978 is 7.4 times larger than thatfor 1968.

b. The rate of increase for 1998 is 7.4 times larger than that for 1988.

c. Answers may vary. For example,data from the past are not necessarilygood indicators of what will happenin the future. Interest rates change,borrowing may decrease, principalmay be paid off early.

8.

9.

From the graph, the values of quickly rise in the range of about

The slope for thesevalues is positive and steep. Then asthe graph nears , the steepnessof the slope decreases and seems toget very close to 0. One can reasonthat the car quickly accelerates for thefirst 20 units of time. Then, it seemsto maintain a constant acceleration forthe rest of the time. To verify this, onecould differentiate and look at valueswhere is increasing.

Section 5.3, pp. 245–247

1. a. absolute max: about 0.3849,absolute min: 0

b. absolute max: about 10.043,absolute min: about

2. a. max: 0.3849,min: 0;

max: about 10,min: about

b. The graphing approach seems to beeasier to use for the functions. It isquicker and it gives the graphs ofthe functions in a good viewingrectangle. The only problem maycome in the second function,because for , the functionquickly approaches values in thenegative thousands.

3. a. 500 squirrelsb. 2000 squirrelsc. (54.9, 10)d.

e. P grows exponentially until thepoint of inflection, then the growthrate decreases and the curvebecomes concave down.

4. a. 1001 itemsb. 500 items

5. 500 units6. 47.2%; 0.462 h7. a.

b. The growth rate of capitalinvestment grew from 468 milliondollars per year in 1947 to 2.112billion dollars per year in 1967.

c. 7.5%d. dollars,

dollars year

e. Statistics Canada data shows theactual amount of U.S. investment in1977 was dollars. Theerror in the model is 3.5%.

f. dollars,

dollars year8. a. 478 158; 38.2 min after the drug was

introducedb. 42.72 min after the drug was

introduced9. 10 h of study should be assigned to the

first exam and 20 h of study for thesecond exam.

10. Use the algorithm for finding extremevalues. First, find the derivate .Then find any critical points by setting

and solving for x. Also findthe values of x for which isundefined. Together these are thecritical values. Now evaluate forthe critical values and the endpoints 2and The highest value will be theabsolute maximum on the interval, andthe lowest value will be the absoluteminimum on the interval.

11. a. is increasing on the intervals and

Also, is decreasing on theinterval

b. 01�2, 0 2 .f 1x 2 10, q 2 .1�q, �2 2f 1x 2

�2.

f 1x 2f ¿ 1x 2f ¿ 1x 2 � 0

f ¿ 1x 2

>dCdt � 42.975 � 109

C � 570.490 � 109

62.5 � 109

>dC

dt� 4.4849 � 109

C � 59.537 � 109

16

14

12

10

8

6

4

2

0 20 40 60 80 100Years since 1867

Cap

ital

inve

stm

ent

from

U.S

. sou

rces

($10

0 m

illio

n)

t

C(t)

25

20

15

10

5

0 50 100 150 200 250 300

P(t)

t

x 6 1.5m1x 2 ,

�5961m1x 2 :f 1x 2 : �5961.916

v¿ 1t 2

t � 20

0 � t � 15.

v1t 2

120

100

80

60

40

20

0 20 40 60 80 100 120

t

v(t)

y

x

1

2

0

–1

–2

–4 –2–6–8 4 6 82

y � 1

>>>�9.12

3.80�23.03x � y � 13.03 � 0�16.64x � y � 25.92 � 0

�3 ln 10

4

3x2 3x ln 3 � 4 4

2x3

�2t

t2�

2t ln 2

t

13 2x2 3 12x2 ln 3 2 � 1 45x 3 1x5 � ln 5 2 � 5x4 440012 2x�3ln 2

2x13x2�2 2 ln 3

1105�6n�n2 2 ln 101�6 � 2n 231103t�5 2 ln 10

ln 3.113.1 2 x � 3x2

3123x 2 ln 2

15! � p .� 1

4! �� 13!

� 12!e1 � e � 1 � 1

1!

ex � e1 � ex � 1.

� x3

3! � x4

4! � p .

ex � 1 � x1

1! � x2

2!


12. a. no maximum or minimum value

b. min: no max

c. min: no max

d. no max or min

13. about 0.6114. a.

b. increasing when and

decreasing when

c. about sd.

15. The solution starts in a similar way to that of question 9. The effectivenessfunction is

The derivative simplifies to

This expression is very difficult tosolve analytically. By calculation on agraphing calculator, we can determinethat the maximum effectiveness occurswhen h.

16. a.

b. after 4.6 days, 5012c. The rate of growth is slowing down

as the colony is getting closer to itslimiting value.

Mid-Chapter Review, pp. 248–249

1. a.b.c.d.e.

f.

2. a.b.

3.4. a. ,

b. ,

c. ,

5. a.b.c.d.e.f.

6. a. 5500

b.

c. decreasing by about 15 rabbits/monthd. 5500

e.

The graph is constantly decreasing.The y-intercept is (0, 5500). Rabbitpopulations normally growexponentially, but this population isshrinking exponentially. Perhaps alarge number of rabbit predators,such as snakes, recently began toappear in the forest. A large numberof predators would quickly shrinkthe rabbit population.

7. at about 0.41 h8. The original function represents

growth when meaning that c and k must have the same sign. Theoriginal function represents decaywhen c and k have opposite signs.

9. a. 5000b. 5751c. 9111

10. a. 406.80 mm Hgb. 316.82 mm Hgc. 246.74 mm Hg

11. ; 15% per year12.

So,

This means that the function is increasing when

13.

14. a.b.c. ,

,

d. NoA¿ 110 2 � $104.35A¿ 15 2 � $77.98A¿ 12 2 � $65.47A¿ 1t 2 � 100011.06 2 t ln 1.06A1t 2 � 100011.06 2 t

0

–1

–2

1

2

1 2–1–2

y

x

x 7 �1.

x 7 �1x � 1 7 0

ex 7 0� ex1x � 1 2f ¿ 1x 2 � xex � 11 2ex

f 1x 2 � xexA � �30e�0.3x

ck 7 0,

6000

4000

2000

0100 20 30 40

�50Qe� t10R

4x1 1ln 4 2 1x � 2 22 � 2x � 4 21ln 1.9 2 11.9 2 x � 1.9x0.99001ln 5 2 15 23x�12x1 1ln 2 2 1x2 2 � 2x 20.641ln 10 2 1 110 20.2x 221ln 8 2 182x�5 2y– � 2ex � xexy¿ � 3ex � xexy– � 4xe2x � 4e2xy¿ � 2xe2x � e2xy– � �3exy¿ � �3ex

x � y � 2 � 0�2.5�500e�5t

41ex � e�x 2221x � xe�x � e�x � e�2x 21ex 2 1x2 � 1 2e�2x1�2x3 � 3x2 2e

17x

�15e�3x

10

8

6

4

2

0 2 4 6 8 10 12Days

Num

ber o

f cel

ls (t

hous

ands

)

t

P

t � 8.16

� 0.03e�25�t10 15 � t 2E¿ 1t 2 � 0.05e� t

10 110 � t 2� 0.6Q9 � 125 � t 2e�25�t

20 RE1t 2 � 0.5Q10 � te� t

10R

t 7 10 s>106.15°

t 7 1ln 2

0 6 t 6 1ln 2

120

100

80

60

40

20

0 2 4 6 8 10

t

d(t)

4

6

8

2

–4

–2

–6

–8

–4 –2–6–8 4 6 820

y

x

4

6

8

2

–2–4 –2–6–8 20

y

x

�e�1 � �0.37,

4

6

8

2

–2–4 –2–6–8 4 6 820

y

x

�e�1 � 3 � 2.63,

–6–8 6 80

4

2

–2

6

8

y

x

2 4–4 –2


e.

f. All the ratios are equivalent (theyequal which is about 0.058 27), which means that

is constant.

15.

Section 5.4, pp. 256–257

1. a.b.c.d.e.f.g.h.i.

j.

2. a.

b.

c.

d.

e.f.

3. a.

b.c.

d.

e.

f.4. a. One could easily find and

to see that they both equalHowever, it is

easier to notice a fundamentaltrigonometric identity. It is knownthat So,

Therefore, is in fact equal toSo, because

b. and are each others’negative. That is,

, while

5. a.

b.

c.

d.

6. a. absolute max: ,absolute min:

b. absolute max: 2.26,absolute min:

c. absolute max: ,absolute min:

d. absolute max: 5,absolute min:

7. a. for positiveintegers k

b. 168. a.

b.

9. ,

10.11. a. for positive

integers kb. 16c. minimum: 0 maximum: 4

12.

13.

14. First find

So,

Therefore,

Section 5.5, p. 260

1. a.b.c.

d.

e.f.

2. a.

b.3. a.

b.c.d.e.

f.

4. a.

b.5. and

6.

7.

The denominator is never negative.for since

reaches its minimum of at

Since the derivative of the

original function is always positive inthe specified interval, the function isalways increasing in that interval.

8.

9. Write tan and use the quotient rule to derive the derivative of the tangent function.

10.11. f – 1x 2 � 32 csc2 14x 2 cot 14x 2�csc2 x

x �sin xcos x

�4x � y � 12 � p 2 � 0

x � �p2 .

�1sin x

�p2 6 x 6 p2 ,1 � sin x 7 0

�1 � sin x

cos2 x

�cos2 x � sin x � sin2 x

cos2 x

�cos2 x � 1�sin x � sin2 x 2

cos2 x

dy

dx�

cos2 x � 11 � sin x 2 1�sin x 2cos2 x

�1 � sin x

cos x

�1

cos x�

sin x

cos x

y � sec x � tan x

ap4

, 0.57 b2pp,x � 0,

2 sec2 x11 � 3 tan2 x 2cos x � sec x �

2 sin2 x

cos 3 x

1

2�xetan�x sec2�x

sin2 x13 tan x cos x � sin x sec2 x 221tan x � cos x 2 1sec2 x � sin x 2�2 tan 1cos x 2sec2 1cos x 2sin x�4x 3 tan1x2 � 1 2 4�3sec2 1x2 � 1 2cos x sec2 1sin x 2y � 4x

y � 2x �p

2� 1

151tan 5x cos 5x � sin 5x sec2 5x 22x sec2 1x2 2 � 2 tan x sec2 x

x12 tan px � px sec2 px 2tan2 px

6x2 tan 1x3 2sec21x3 22 sec2 x � 2 sec2

2x3 sec2 3x

y– � k2y � 0.

� 0� k2A cos kt � k2B sin kt

� �k2A cos kt � k2B sin kt

� k21A cos kt � B sin kt 2� �k2A cos kt � k2B sin kt

y– � k2y

y– � �k2A cos kt � k2B sin kt

y¿ � �kA sin kt � kB cos kty � A cos kt � B sin kt

y–.u �p

6

u �p

3

3p8 � pkt �

p8 � pk,

��3

1sec x 2 ¿ � sec x tan x1csc x 2 ¿ � �csc x cot xap

4,�2 b

p 2p

2

1

0–1

–2

x

f (x)

3p4 � pkt �

p4 � pk,

�5

��2�2

�5.14

��2�2

� 12x � 2 sin x cos x 2m¿ 1x 2 � 31x2 � cos2 x 22� sin 2x sin 3x cos x� 2 sin x sin 3x cos 2x

� 3 sin x sin 2x cos 3xh¿ 1x 2v¿ 1t 2 �

�sin t � 2 1sin t 2 1cos t 22�1 � cos t � sin2 t

v¿ 1t 2 �sin 1�t 2 cos 1�t 2

�t

g¿ 1x 2 � �21sin x 2 1cos x 2 .f ¿ 1x 2 � 1sin x 2 1cos x 2g¿ 1x 2f ¿ 1x 2

f ¿ 1x 2 � g¿ 1x 2 . f 1x 2 � g1x 2 ,g1x 2 . f 1x 2sin2 x � 1 � cos2 x.sin2 x � cos2 x � 1.

2 1sin x 2 1cos x 2 .g¿ 1x 2 f ¿ 1x 22x � y � p � 0

y ��3

2� � a x �

p

4b

y � �3 a x �p

2b

y � �1�2x � y � 0

�x � 2y � ap3

� �3 b � 0

� 3x sin x � 3 cos x2x3 cos x � 6x2 sin xex12 cos x 2

1

1 � cos x

�sin 1sin 2x 2 � 2 cos 2x

�2 sin 2x

x�

cos 2x

x2

2 cos 12x 2�

1

x2 cos a 1

xb

2x � sin x9 cos 13x � 2p 2ex cos 1ex 22x1ln 2 2 � 2 cos x � 2 sin x3 cos 13x 2 � 4 sin 14x 28 sin 1�4x 213x2 � 2 2 1cos 1x3 � 2x � 4 2 2�6 sin 3x2 cos 2x

y � y¿ � cex� cex

y¿ � c1ex 2 � 10 2 1ex 2y � cex

A¿ 1t 2A1t 2

ln 1.06,

A¿ 110 2A110 2 � ln 1.06

A¿ 15 2A15 2 � ln 1.06,

A¿ 12 2A12 2 � ln 1.06,



1. a.b.c.

d.e.

f.

2. a.b.c.d.

e.

f.

3. a.b.

c.

d.e.f.

4. a.b. The function has a horizontal

tangent at (1, e). So this point couldbe possible local max or min.

5. a.b. The slope of the tangent to at

the point with x-coordinate is 0.

6. a.b.

7.

Now,

8.9.

10. about m per unit of time

11. a.b. After 10 days, about 0.1156 mice

are infected per day. Essentially,almost 0 mice are infected per daywhen

12. a.b.

13. a.b.c.d.

14. a.b.

c.

d.e.f.

15. a.b.

c.

d.e.f.

16.

17.Thus,

The acceleration at any time t is

Hence,Now,

18. displacement: 5,velocity: 10,acceleration: 20

19. each angle rad, or 20. 4.5 m21. 2.5 m22. 5.19 ft23. a.

b.


1. a.b.

c.

d.

e.

f.

2. ,The tangent line is the given line.

3.4. a.

Thus, the acceleration is a constantmultiple of the velocity. As thevelocity of the particle decreases,the acceleration increases by afactor of k.

b. 10 cm s

c.

5. a.b.

6. absolute max: 1,absolute min: 0

7. 40.24

8. minimum: , no maximum

9. a.

b. increasing:

decreasing: and

c. local maximum at

local minimum at

d.

Cumulative Review of Calculus,pp. 267–270

1. a. 16 c.

b. d.2. a. m s

b. m s3.4. a. m s

b. m sc. m s>53.655

>19.6>19.6

f 1x 2 � x3>15>13

160 ln 2�2

1

6

–p p3p4

0

2

1

–2

–1

–3

–4

3

4

y

x– 3p4

p2– p

2p4– p

4

1�5p6 , �2.598 21�p6 , 2.598 2�

p6 6 x 6 p

�p � x 6 �5p6

�5p6 6 x 6 �

p6 ;

x � �p

6, �

5p

6,p

2

Q� 1e4R

� csc3 x � sin xf – 1x 2 � csc x cot2 xf – 1x 2 � 21sin2 x � cos2 x 2ln 2

k; �5k

>

� �kv1t 2� �k110e�kt 2a1t 2 � v¿ 1t 2 � �10ke�kt�2x � y � 1

�6x � y � 2

�sec2

�1 � x

2�1 � x

6x sin2 1x2 2cos 1x2 2

2 cos x � 15 sin 5x

3

23e3x � e�3x 4

3x2�3x � ln 3 � 12x � 3 2�4xe�2x2

f – 1x 2 � 2 sec3 x � 2 sec x tan2 x� 8 cos2 1x � 2 2f – 1x 2 � �8 sin2 1x � 2 2

45°p4

� 100p218 sin 110pt 22 � 0.

d2sdt2 � 100p2s � �800p2 sin 110pt 2� �800p2sin 110pt 2 .a � 80p1�sin 110pt 22 110p 2a �

dvdt �

d2sdt2.

� 80p cos 110pt 2v � 81cos 110pt 22 110p 2v �dsdt;

x � y �p

2� 0

2 sin x cos2 x � sin3 x�2 cos x sin xcos2 x � sin2 x

�cos ap2

� x bx2 cos x � 2x sin x2x ln 2 cos 2x

�6110 23x ln 104ex512 2x ln 2

2152 22x ln 52

10.47 2 x ln 10.47 25x ln 5

�25e5x11 � e5x 24ex�exexe�1�9e�x12 � 3e�x 22c1

c2

t � 10.

t � 20

0.3928

�x � y � 03x � y � 2 ln 2 � 2 � 0

�4e2x

132x � 1 22 �dy

dx

�e4x � 2e2x � 1 � e4x � 2e2x � 11e2x � 1 22

1 � y2 � 1 �e4x � 2e2x � 11e2x � 1 22

�4e2x

1e2x � 1 22�

2e4x � 2e2x � 2e4x � 2e2x

1e2x � 1 22dy

dx�

2e2x1e2x � 1 2 � 1e2x � 1 2 12e2x 21e2x � 1 22y �

e2x � 1

e2x � 1

20e10x15x � 1 2ex1x � 1 212

f 1x 20

x � 1�4 cos 12x 2sin 12x 2e3x13 sin 2x � 2 cos 2x 22x sec2 12x 2 � tan 2x

�sin x12 � cos x 22

3 sec2 13x 26 cos 12x 2 � 8 sin 12x 25�x a�

1

x2�

ln 5

2x�xb

4 � 4x ln 4

4x

x3 � 2x1x ln 2 � 4 25 � 5x1x ln 5 � 1 26x143x2 2 ln 410x ln 10

2et

1et � 1 22ex1x � 1 21�6x � 5 2e�3x2�5x

2e2x�32 � 3ex�ex


5. a. 19 000 fish yearb. 23 000 fish year

6. a. i. 3ii. 1

iii. 3iv. 2

b. No, does not exist. In order

for the limit to exist,

and must exist and they

must be the same. In this case,but

so

does not exist.7. is discontinuous at

but

8. a. d.

b. 6 e.

c. f.

9. a.

b.

10. a.

b.

c.

d.

e.

f.

11.12. 313. a.

b. 46 people per yearc. 2006

14. a.

b.

c.

d.

15. a. maximum: 82, minimum: 6b. maximum: minimum: 2

c. maximum: minimum:

d. maximum: 5, minimum: 1

16. a.

b. stationary when or advancing when andretreating when

c.d.e.

17.18. cm, cm19. cm, cm20. a.

b. 46.7 cm by 46.7 cmby 46.6 cm

21.22. $70 or $8023. $1140

24. a.

is critical number,Increase:Decrease:

b.

is critical number,

Increase:

Decrease:

c.

are critical numbers,Increase:Decrease:

d. The function has

no critical numbers. The function is decreasing everywhere it is defined,that is,

25. a. is a horizontal asymptote.are the vertical asymptotes.

There is no oblique asymptote.

is a local maximum.

b. There are no horizontal asymptotes.are the vertical asymptotes.

is an oblique asymptote.is a local

maximum, is a localminimum.

26. a.

b.

27. a.b.c.d.

28.29. a. 5 days

b. 2730. a.

b.

c.

d.

e.f.

31. about 4.8 m32. about 8.5 m

Chapter 6

Review of Prerequisite Skills,p. 273

1. a. d.

b. e.

c. f. 1

2.

3. a. , ,

b. , ,

4. , ,5. , ,6. 5.82 km7. 8.66 km8. 21.1 km9. 59.4 cm2

�T � 34°�S � 102°�R � 44°YZ � 6.78XZ � 7.36�Z � 50°

�C � 52.4°�B � 29.7°�A � 97.9°

�C � 53.5°�B � 36.5°AB � 29.7

4

3

1

2

�2

2��3

�3

2

�3

2

�2x sin x2 cos 1cos x2 22x sec2 x2 � 2 tan x sec2 x

1 � 2 cos x1cos x � 2 222x � 3 cos 3x

2�x2 � sin 3x

8 cos 2x 1sin 2x � 1 232 cos x � 15 sin 5x

y � 2e1x � 1 2 � e1cos x 2esin x13 ln 6 263x�8e3x13x � 1 21�20 2e5x�1

4 6

3x——–—–x2 – 4

y =

0

4

6

2

–4

–6

–2

y

x

2–2–4–6

2

y = 4x3 + 6x2 – 24x – 2

0

20

30

10

–10

y

x

1–1–2–3

1�3, 6�3 21��3, �6�3 2y � 4xx � ;1

Q0, �89R

x � ;3y � 0

x � 2.

dydx � �

21x � 2 2 2.�2 6 x 6 2

x 7 2,x 6 �2,x � ;2

dy

dx� 6x2 � 24,

x 6 �43

x 7 �43,

x � �43

dy

dx� 12x � 16,

x 7 2x 6 2,

x � 2

dy

dx� �10x � 20,

x � 4

101 629.5 cm3;140 � 2x

h � 27.5r � 6.8h � 8.6r � 4.3

14 062.5 m24.5 6 t � 80 � t 6 4.5t � 4.5

v1t 2 6 0v1t 2 7 0,

t � 3,t � 6a1t 2 � 18t � 81v1t 2 � 9t2 � 81t � 162,

12

e4

1 � e4 ,

9 13,

f – 1x 2 � 12x2 �20

x6

f ¿ 1x 2 � 4x3 �4

x5;

f – 1x 2 �3

�x5f ¿ 1x 2 � �

2

�x3;

f – 1x 2 � �12

x4f ¿ 1x 2 �

4

x3;

f – 1x 2 � 20x3 � 30xf ¿ 1x 2 � 5x4 � 15x2 � 1;

p¿ 1t 2 � 4t � 6

4x � 3y � 10 � 0� 32x � 612x � 1 22 45 3x2 � 12x � 1 23 4414x2 � 1 24184x2 � 80x � 9 213x � 2 24� 1x2 � 3 22120x4 � 5 24x1x2 � 3 2 14x5 � 5x � 1 2

61x � 3 223x2

�2x3 � 1

3x2 � 8x � 5

�1

x2

6x � 1

1

2�

1

9

1

12

4

3�

1

5

limxS2�

f 1x 2 � 3.limxS2�

f 1x 2 � 5,x � 2.f 1x 2limxS4

f 1x 2limxS4�

f 1x 2 � �q,

limxS4�

f 1x 2 �q,

limxS4�

f 1x 2 limxS4�

f 1x 2limxS4

f 1x 2

>>


Section 6.1, pp. 279–281

1. a. False; two vectors with the samemagnitude can have differentdirections, so they are not equal.

b. True; equal vectors have the samedirection and the same magnitude.

c. False; equal or opposite vectorsmust be parallel and have the samemagnitude. If two parallel vectorshave different magnitude, theycannot be equal or opposite.

d. False; equal or opposite vectorsmust be parallel and have the samemagnitude. Two vectors with thesame magnitude can have directionsthat are not parallel, so they are notequal or opposite.

2. The following are scalars: height,temperature, mass, area, volume,distance, and speed. There is not adirection associated with any of thesequantities.The following are vectors: weight,displacement, force, and velocity.There is a direction associated witheach of these quantities.

3. Answers may vary. For example:A rolling ball stops due to friction,which resists the direction of motion.A swinging pendulum stops due tofriction resisting the swingingpendulum.

4. Answers may vary. For example:a.

b.

c. & & 5.

a.b.c. but

d.e.

6. Drawings not to scale.

a. b. c.

d. e.

7. a. 100 km h, southb. 50 km h, westc. 100 km h, northeastd. 25 km h, northweste. 60 km h, east

8. a. 400 km h, due southb. 70 km h, southwesterlyc. 30 km h southeasterlyd. 25 km h, due east

9. a. i. False; they have equal magnitude,but opposite direction.

ii. True; they have equal magnitude.iii. True; the base has sides of equal

length, so the vectors have equalmagnitude.

iv. True; they have equal magnitudeand direction.

b. , ,

10. a. The tangent vector describesJames’s velocity at that moment. At point A, his speed is 15 km h andhe is heading north. The tangentvector shows his velocity is 15 km h, north.

b. James’s speedc. The magnitude of James’s velocity

(his speed) is constant, but thedirection of his velocity changes atevery point.

d. Ce. 3.5 minf. southwest

11. a. or 3.16b.c.d. (0, 0)

Section 6.2, pp. 290–292

1. a.

b.

c.

d.

2. a.

b.

c.

d.

CA!

A

C B

CB!

A

C B

0!

A

C B

BA!

A

C B

–x–y

–y –x

–x

y –xy

x

x –y–y

x

x + y

y

10, �3 21�3, 1 2�10

>>

@BH! @ � �82

@BE! @ � �73@BD

! @ � �18

>>>>>>>>>

1.5 cm3 cm

4 cm

AB!� �2JI

!GH!� 2AB

!AB!� EF

!�AB!� � �EF

!�

AB!� �EF

!AB!� CD

!

B

A

D

C

J

I

H

G

E

F

EB!

AE!

DB!;AC

! ED!� �EB

!AE!

� �CE!;

AB!� �CD

!;AD

!� �CB

!;

DE!� EB

! AE!� EC

!;AB

!� DC

!;AD

!� BC

!;


10 cm

6 cm

3. a.

b.

c.

d.

4. a.

b.

c. The resultant vectors are the same.The order in which you add vectorsdoes not matter.

5. a.

b.

6.

so

7. a.

b.

c.

d.

e.

f.

g.

h.8. a.

b. See the figure in part a. for thedrawn vectors.

andso

9. a. 11 km hb.

c. 3 km h

10. a.

b. The vectors form a trianglewith side lengths , and

Find using the

cosine law.

11. 170 km h, N28.1°W12. ,13. 0.5214. The diagonals of a parallelogram

bisect each other. So,

Therefore,

15. Multiple applications of the TriangleLaw for adding vectors show that

(since both are equal to the undrawn vector ),and that (since both are equal to the undrawnvector ).Adding these two equations gives

16. and represent thediagonals of a parallelogram with sides

and

Since and the onlyparallelogram with equal diagonals is arectangle, the parallelogram must alsobe a rectangle.

@ a!� b! @ � @ a!� b

! @ ,b!.a

!

a!� b!

a!� b!

2RM!� TP

!� SQ

!� a!� b!� TP

!� SQ

!2RM!� a!� b!

RQ!

RM!� a!� b!� SQ

! TM!RM

!� b!� a!� TP

!

EA!� EB

!� EC

!� ED

!� 0!.

and ED!� �EB

!.

EA!� �EC

!

u � 73.7°0x!� y! 0 � 25>� V @ f1! @ 2 � @ f2! @ 2 � 2 @ f1! @ @ f2! @ cos 1u 2@ f1!� f2! @ � 2 @ f1! @ @ f2! @ cos 1u 2@ f1! @ 2 � @ f2! @ 2@ f1!� f2! @ 2 �

@ f1!� f2! @@ f1!� f2

! @ . @ f2! @@ f1! @ ,

f1 + f2

u f1

f2

ship

4 km/h

7 km/h

3 km/h

>

4 km/h

7 km/h

11 km/h

> � 0x!� y! 0 20y!� x

! 0 2 � 0x! 0 ,0�x! 02 0y! 0 0�x

! 0 cos 1u 2� 0y! 0 2 � 0x! 0 2 �0y!� x! 0 2

B Cu

u

DA

y

y

y

y

xxx –

xy –

�x!� z!

�x!� y!� z!

�x!� y!

x!� y!� z!

�x!� y!

x!� y!

y!

�x! � MQ

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!2 � MS!� SQ

!

� SQ!z

!� t!� ST

!� TQ

!� MS!x

!� y!� MR

!� RS

!

RQ

–RS

–PQ

Q

R

P

S

PS

0!

–RQ QR

P

SPQ

PS

RS

PR

� PS!

� a!� 1b!� c

!21a!� b!2 � c

!

(a + b)

(a + b) + c

b

ca

(b + c)

a + (b + c)

b

ca

a – b + c

c

–b

a

a – b – c

–c

–b

a

a + b – c

–c

b

a

a + b + c

c

b

a


17.

Let point M be defined as shown. Twoapplications of the triangle law foradding vectors show that

Adding these two equations gives

From the given information,and

(since they areopposing vectors of equal length), so

, as desired.

Section 6.3, pp. 298–301

1. A vector cannot equal a scalar.2. Drawings not to scale.

a.

b.

c.

d.

3. E25°N describes a direction that is 25°toward the north of due east. N65°E and“a bearing of 65°” both describe adirection that is 65° toward the east ofdue north.

4. Drawings not to scale. Answers mayvary. For example:a.

z

c.

d.

e.

5. Drawings not to scale.

a.

b.

c.

d.

6. Answers may vary. For example:

a.

b.

c.

d.

e.

2a – 3b

aa

–b–b

–b

2a + 3b

aa b

bb

–3b

3b

2a

a b

–x

–x

–2x – y

–y

y –x

–x

–2x + y

–y

–y

–y

x

x – 3y

y

y

y

x

x + 3y6 cm

2 cm

9 cm

3 cm

GQ!� GP

!� GR

!� 0!

QM!� RM

!� 0!2MG

!� GP

!

� GR!� RM

!� 0!

GQ!� QM

!� 2 MG

!

GR!� RM

!� MG

!� 0!GQ

!� QM

!� MG

!� 0!

G

RQ

P

M


2v!

1

2v!

�2

3 v!

�2v!

10v! 0 v!

7. a. Answers may vary. For example:infinitely many

b. Answers may vary. For example:not unique

8.

9. yes

10. a. collinearb. not collinearc. not collineard. collinear

11. a. is a vector with length 1 unit in

the same direction as

b. is a vector with length 1 unit

in the opposite direction of

12.

13. a. d.

b. e.

c.

14. or 2.24, from toward

15. 2.91, 9.9° from toward

16.

17.

But

So, or

18.

Notice that

We can conclude that is parallel to

and

19.

Answers may vary. For example:

a. andb. and

c. andd. and

20. a.b.

21. a.

b.

The two are, therefore, parallel

(collinear) and .22. Applying the triangle law for adding

vectors shows that

The given information states that

By the properties of trapezoids, thisgives

, and since

, the originalequation gives

Section 6.4, pp. 306–307

1. a. 0b. 1c.d. 1

2.

3.

4. Answers may vary. For example:

5.

6. a.

b.c. Yes, the diagonals of a rectangular

prism are of equal length.

7.

8. a.

b.

c.

9. ,

10.

� �1

3b!

�2

3b!� b!

�2

31y!� z

!2 � b!

�2

3y!�

2

3z!� b!

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3y!�

1

3z!� 1b!� z

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y!�

1

13a!�

12

13b!

x!�

5

13a!�

18

13b!

� �6i!� 13j

!� 7k!�7i

!� 11j

!� 4k!12i

!� 17j

!� 5k!�4a

!� 9b

!� 24c

!

0!EC!

� PQ!

� PS!� SQ

!� SQ

!� PS

!� 1SR

!� RQ

!2 � 1PT!� TS

!2� 1RQ!� SR

!2 � 1TS!� PT

!2� TS!� PT

!PQ!� RQ

!� SR

!

k(a + b)a + b

a

b

k ak b

(b + c)

(a + b)a

c

b

(a + b)+ c

a + b + c

a + (b+ c)==

a!� b!� b!� a!

a + b

b + abb

a

a

0!

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2

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3

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3

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3

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!

3

2AE!� EC

!

3

2AB!� DC

!

AB!�

2

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BE!� 2 @CD

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! v!� DB

!u!� AC

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A B

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DC

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12 b!.AD

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!0b! 0 � 1

0b! 0 � 10a! 0 0a! 00b! 0 � ` 10a! 0 a! `b!�

10a! 0 a!y!

x!y

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u � 26.6°�5

1

30a! 0

4

3a!1

3a!

2

30a! 0�

2

3a!

m �2

3

x!.

�10x! 0 x!

x!.

10x! 0 x!

b

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4aa

b!

a!

d � 2, e � 0, f � �1;

n � �3,m � 4,


a a

bb b

11. a.

b.

12. Applying the triangle law for addingvectors shows that

The given information states that

By the properties of trapezoids, thisgives

, and since

, the original equationgives

Mid-Chapter Review, pp. 308–309

1. a. , ,,

There is not enough information todetermine if there is a vector equalto

b.

(parallelogram)

2. a. c. e.b. d. f.

3. a.b. 53°

4. 4 or 5. In quadrilateral PQRS, look at

PQR. Joining the midpoints B and C creates a vector that isparallel to and half the length of

. Look at SPR. Joining themidpoints A and D creates a vector

that is parallel to and halfthe length of . is parallel to

and equal in length to .Therefore, ABCD is aparallelogram.

6. a.b. 71°

c.

d.7. 38.

9. , ,

,10. Construct a parallelogram with sides

and . Since the diagonalsbisect each other, 2 is the diagonalequal to . Or

and

So, .

And .

Now ,

Multiplying by 2 gives

11.

12. 460 km h, south

13. a.b.c.

14. a.

b.

c.

d.

15.

Section 6.5, pp. 316–318

1. No, as the y-coordinate is not a realnumber.

2. a. We first arrange the x-, y-, and z-axes (each a copy of the real line)in a way so that each pair of axes are perpendicular to each other (i.e.,the x- and y-axes are arranged intheir usual way to form the xy-plane, and the z-axis passes throughthe origin of the xy-plane and isperpendicular to this plane). This iseasiest viewed as a “right-handedsystem,” where, from the viewer’sperspective, the positive z-axispoints upward, the positive x-axispoints out of the page, and thepositive y-axis points rightward inthe plane of the page. Then, givenpoint P(a, b, c), we locate thispoint’s unique position by moving a units along the x-axis, then fromthere b units parallel to the y-axis,and finally c units parallel to the z-axis. It’s associated uniqueposition vector is determined bydrawing a vector with tail at theorigin O(0, 0, 0) and head at P.

b. Since this position vector is unique,its coordinates are unique.Therefore, and

3. a. andb.

4. This is not an acceptable vector in asthe z-coordinate is not an integer.However, since all of the coordinatesare real numbers, this is acceptable as avector in

5.

y

z

x

(0, –4, 0)

(0, 0, –2)

(0, –4, –2)

(4, 0, 0)

(4, –4, 0)

(4, 0, –2)A(4, –4, –2)

O(0, 0, 0)

R3.

I3

15, �3, 8 2 c � 8.b � �3,a � 5,c � �8.

b � �3,a � �4,

� �3a!RS

!� QS

!� QR

!� 3b!� a!

PS!� PQ

!� QS

!

a12

b12

a

–b

–2 b

a32

b

a13 a + b1

3

SR!PT!PT! >

� 2x!� y!

BC!BD!� 2x!� y!

2 OB!� OA

!� OC

! .

�12 1OC

!� OA

!2OB! � OA

!

AC!� OC

!� OA

!OB! � OA

! �

1

2AC!

AB!�

1

2AC!.OB

! � OA

! � AB

!

OA! � OC

!

OB!

OC! OA

!

AC!� x!� y!

BD!� �x

!� y!

DC!� x!

BC!� �y

!0m!� n

! 0 � @ 0m! 0 � 0n! 0 @20�7

10u!� v! 0 1u!� v

!2 �1

2�211u!� v

!22�21

AD!

AD! BC

!PR! PR

!AD!

^PR! PR

! BC!^

t � �4t �

�3

PQ!

RU!

RV! PS

!PS!

RV!

� @BC! @@PD

! @ � @DA! @AP

!.

CB!� DA

!AD!� BC

! BA!� CD

!AB!� DC

!

TO!�

2

3TZ!�

1

3TX!

3

2TO!� TZ

!�

1

2TX!

TO!�

1

2TO!� TZ

!�

1

2TX!

TY!� TO

!� OY

!

1

2TO!� OY

!

1

2TX!� ZY

!TX!� 2ZY

!TY!� TZ

!� ZY

!

�AG!� � �BH

!��

� �BH!�2

� 0�a! 0 2 � 0b! 0 2 � 0c! 0 2�AG

!�2 � 0a! 0 2 � 0b! 0 2 � 0c! 0 2DF!� a!� b!� c!

CE!� �a

!� b!� c!,

BH!� �a

!� b!� c!,

AG!� a!� b!� c!,


6. a. is located on the y-axis.and are three otherpoints on this axis.

b. the vector withtail at the origin and head at A.

7. a. Answers may vary. For example:

b. Yes, these vectors are collinear(parallel), as they all lie on the sameline, in this case the z-axis.

c. A general vector lying on the z-axiswould be of the form for any real number a. Therefore, thisvector would be represented byplacing the tail at O and the head atthe point (0, 0, a) on the z-axis.

8.

9. a.

b.10. a.

b.

c.

d.

e.

f.

11. a.

b.

c.

OCy

z

x

O(0, 0, 0)

(1, 0, 0)

(0, –2, 0)C(1, –2, 1)

(1, 0, 1)

(0, –2, 1)

(1, –2, 0)

(0, 0, 1)

OC!� 11, �2, 1 2

(0, 1, 1)

(–2, 1, 0)

(–2, 0, 1)B(–2, 1, 1)

(0, 1, 0)

O(0, 0, 0)

(0, 0, 1)

(–2, 0, 0)

y

z

x

OB

OB!� 1�2, 1, 1 2

y

z(0, 0, 3)

(1, 0, 3)(0, 2, 3)

(0, 2, 0)

(1, 2, 0)

A(1, 2, 3)

O(0, 0, 0)

(1, 0, 0)

OA

x

OA!� 11, 2, 3 2

y

z

O(0, 0, 0)

F(1, –1, –1)

(0, 0, –1)

(1, –1, 0)

(0, –1, 0)

(1, 0, 0)

(1, 0, –1)

(0, –1, –1)

x

y

z

E(1, –1, 1)

(0, –1, 1)

O(0, 0, 0)

(1, 0, 0)

(0, –1, 0)

(0, 0, 1)

(1, –1, 0)

(1, 0, 1)

x

y

z

(1, 0, 1)

D(1, 1, 1)(0, 0, 1)

(0, 1, 1)

O(0, 0, 0)

(1, 0, 0)

(0, 1, 0)

(1, 1, 0)

x

y

z

O(0, 0, 0)

(1, 0, 0)

(0, –2, 0)C(1, –2, 1)

(1, 0, 1)

(0, –2, 1)

(1, –2, 0)

(0, 0, 1)

(0, 1, 1)

(–2, 1, 0)

(–2, 0, 1)B(–2, 1, 1)

(0, 1, 0)

O(0, 0, 0)

(0, 0, 1)

(–2, 0, 0)

y

z

x

y

z(0, 0, 3)

(1, 0, 3)(0, 2, 3)

(0, 2, 0)

(1, 2, 0)

A(1, 2, 3)

O(0, 0, 0)

(1, 0, 0)

x

z � �4

y

z

x

A(3, 2, –4)B(1, 1, –4)

C(0, 1, –4)

A(1, 0, 0)

B(0, –2, 0)

C(0, 0, –3)

D(2, 3, 0)

E(2, 0, 3)F(0, 2, 3)

y

z

x

OA!� 10, 0, a 2

OC!� 10, 0, �5 2OB

!� 10, 0, �1 2 ,OA

!� 10, 0, 1 2 ,

O10, 0, 0 2OA!� 10, �1, 0 2 ,D10, 10, 0 2 C10, 2, 0 2 ,B10, �2, 0 2A10, �1, 0 2

y

z

x

(0, 4, 0)

(0, 0, –2)(4, 0, 0)

(0, 4, –2)

(4, 0, –2) C(4, 4, –2)(4, 4, 0)

O(0, 0, 0)

(0, 0, 2)

(0, 4, 2)

B(–4, 4, 2)(–4, 0, 2)

(–4, 4, 0)

(–4, 0, 0)

O(0, 0, 0)

(0, 4, 0)

y

z

x


d.

e.

f.

12. a.b. Since P and Q represent the same

point in , they will have the sameassociated position vector, i.e.,

So, since these vectorsare equal, they will certainly haveequal magnitudes, i.e.,

13. xy-plane, ; xz-plane, ; yz-plane, .

14. a. Every point on the plane containingpoints M, N, and P has y-coordinateequal to 0. Therefore, the equationof the plane containing these pointsis (this is just the xz-plane).

b. The plane contains the originO(0, 0, 0), and so since it alsocontains the points M, N, and P aswell, it will contain the position

vectors associated with these pointsjoining O (tail) to the given point(head). That is, the plane

contains the vectors

and15. a.

b.

c. 7 unitsd.e. Every point contained in rectangle

BCEP has y-coordinate equal to 4,and so is of the form (x, 4, z), wherex and z are real numbers such that

and16. a.

b.

c.

d.

e.

f.

17.

18. First, by the trianglelaw of vector addition, where

and are drawn in standardposition (starting from the origin O

(0, 0, 0)), and is drawn starting

from the head of Notice that OA!

OA!.

OB!

OA!

OP! OB

!� 10, 0, �10 2 ,OA

!� 15, �10, 0 2 ,

OP!� OA

!� OB

!

y

z

x

(0, 0, 1) (0, 1, 1)

(1, 1, 1)

(0, 1, 0)

(1, 1, 0)(1, 0, 0)

(1, 0, 1)

O(0, 0, 0)

J(–2, –2, 0)

y

z

x

O(0, 0, 0)

F(0, 0, 5)

y

z

x

O(0, 0, 0)

y

z

x

M(–1, 3, –2)

O(0, 0, 0)

y

z

x

C(2, 4, 5)

O(0, 0, 0)

D(–3, 4)

x

y

O(0, 0)

x

P(4, –2)

y

O(0, 0)

�7 � z � 0.�2 � x � 0

y � 4

OF!� 1�2, 0, �7 2OE!� 10, 4, �7 2 ,OD

!� 10, 0, �7 2 ,OC

!� 10, 4, 0 2 ,OB!� 1�2, 4, 0 2 ,OA!� 1�2, 0, 0 2 ,F 1�2, 0, �7 2E10, 4, �7 2 ,D10, 0, �7 2 ,C10, 4, 0 2 , B1�2, 4, 0 2 ,A1�2, 0, 0 2 ,OP!.

ON!,OM

!,

y � 0

y � 0y � 0

x � 0y � 0z � 0�OP!� � �OQ

!�.

OP!� OQ

!.

R3

c � 5a � 11

y

z

x

O(0, 0, 0)F(1, –1, –1)

(0, –1, 0)

(0, 0, –1)

(1, –1, 0)

(1, 0, 0)

(1, 0, –1)

(0, –1, –1)

OF

OF!� 11, �1, �1 2

y

z

x

E(1, –1, 1)

(0, –1, 1)

O(0, 0, 0)

(1, 0, 0)

(0, –1, 0)

(0, 0, 1)

(1, –1, 0)

(1, 0, 1)

OE

OE!� 11, �1, 1 2

y

z

x

(1, 0, 1)

D(1, 1, 1)(0, 0, 1)(0, 1, 1)

O(0, 0, 0)

(1, 0, 0)

(0, 1, 0)(1, 1, 0)

OD

OD!� 11, 1, 1 2


lies in the xy-plane, and isperpendicular to the xy-plane (so is

perpendicular to ). So,

and form a right triangle and, bythe Pythagorean theorem,

Similarly, by the triangle law of vector addition, where

and andthese three vectors form a righttriangle as well. So,

Obviously and sosubstituting gives

19.

Section 6.6, pp. 324–326

1. a.

b. ,

c. ,

2.

a.

b.

3. 54. a. ,

b. 5.835. a. ,

b. ,

6. a.b.c. (1, 11)

7. a.

b.

c.8. a. about

b. aboutc. aboutd. about

9. a. , ,

,

b. , ,

,

10. a. , ,

b. so obviously

we will have

11. a.

b. , ,

,

c.

Since , thetriangle is a right triangle.

12. a.

b.

c. , ,13. a. ,

b. ,14. a.

b. Because ABCD is a rectangle, wewill have

So, and i.e.,

15. a.

b.

16.

17. a. aboutb. about 35.4°

80.9°

a 9

41,

40

41b

Q a0, �21

4b

P a 21

10, 0 b

C10, 172.y � 17,x � 0y � 9 � 8x � 6 � 6

1x � 6, y � 9 2 � 16, 8 21x, y 2 � 1�6, 9 2 � 18, 112 � 12, 32BC!� AD

!

x

y

A(2, 3)

D(8, 11)

C(x, y)

B(–6, 9)

y � 4.x � �12y � �2x � 7

Z110, 4 2Y14, �8 2X1�6, 12 2

x

yX(–6, 12)

Y(4, –8)

B(7, –2)

C(2, 8)

A(–1, 2)

Z(10, 4)

x

y

A(–1, 2)

B(7, –2)

C(2, 8)

�CB!�2 � �AC

!�2 � �AB

!�2

�AB!�2 � 25

�CB!�2 � 125, �AC

!�2 � 100,

CB!

� 11.18�AC!� � 10

�AB!� � 5AB

!14, 3 2x

y

A(2, 3)

B(6, 6)C(–4, 11)

�OA!� � �BC

!�.

� BC!,OA

!� 16, 3 2BC!� 16, 3 2 BA

!� 1�5, 9 2OC

!� 117, �3 2�GH

!� � 5�EF

!� � 7.21

�CD!� � 4.47�AB

!� � 4.47

� 15, 0 2GH!

� 1�6, 4 2EF! � 12, 4 2CD

!� 14, 2 2AB! 18.38

18.386.714.12

�29i!� 28j

!3i!� 51j

!7i!� 8j!

1�30, 0 21�2, 7 20a!� b! 0 � 28.28

0a!� b! 0 � 100.020b! 0 � 410a! 0 � 61

b � 5a � �3

2OA! and �2OA

!

1

2OA! and �

1

2OA!,

y

x

10

15

20

5

0

–10

–15

–20

–5–6 –3–9–12 6 9 123

(12, 20)

O(0, 0)

(–3, –5)

(–12, –20)

(3, 5)

y

x

20

25

30

15

10

0–5

5

–6 –3–9 6 93

O(0, 0) OA

A(6, 10)

�BA!� � 3.61�AB

!� � 3.61

�OB!� � 5.39�OA

!� � 3.16

x

B(2, 5)

y

A(–1, 3)

AB

BA

BA!� 1�3, �2 2AB

!� 13, 2 2 ,

16, �5, 2 2� 15�OP!� � �225

� 225� 125 � 100

�OP!�2 � �OA

!�2 � �OB

!�2

�OB!�2 � 100,

� 125� 25 � 100

�OA!�2 � 0a! 0 2 � �b

!�2

b!� 10, �10, 0 2 ,a

!� 15, 0, 0 2

OA!� a!� b!

�OP!�2 � �OA

!�2 � �OB

!�2

OB! OA

!,OP

!,OA

!

OB!


Section 6.7, pp. 332–333

1. a.b. about

2. ,

3. 34. a.

b. , ,

c. ,

,

represents the vector from the

tip of to the tip of It is thedifference between the two vectors.

5. a.b.

c.

d.

6. a.

b.

c.

d.7. a. about

b. aboutc. aboutd. about

8. ,

9. a. The vectors , and represent the xy-plane, xz-plane,and yz-plane, respectively. They are also the vector from the originto points , and

, respectively.

b. ,

,

c. ,

,

d. , is a direction vectorfrom A to B.

10. a. 7b. 13c. (5, 2, 9)d. 10.49e.f. 10.49

11. In order to show that ABCD is aparallelogram, we must show that

or This willshow they have the same direction,thus the opposite sides are parallel.

We have shown and

so ABCD is a parallelogram.

12. , ,

13. a.

b. ,,

,,,,

,

14. (1, 0, 0)

15. 4.36

Section 6.8, pp. 340–341

1. They are collinear, thus a linearcombination is not applicable.

2. It is not possible to use in a spanningset. Therefore, the remaining vectorsonly span

3. The set of vectors spanned by (0, 1) is If we let then

4. spans the set This is anyvector along the x-axis. Examples:(2, 0, 0), .

5. As in question 2, it isn’t possible to usein a spanning set.

6.are all the possible

spanning sets for with 2 vectors.7. a.

b.8.

9. a. It is the set of vectors in the xy-plane.

b.c. By part a., the vector is not in the

xy-plane. There is no combinationthat would produce a number otherthan 0 for the z-component.

d. It would still only span the xy-plane. There would be no need for that vector.

10. , ,11.

12. a. ,

b.

13. a. The statement adoes

not have a consistent solution.b.

14.15. ; Non-parallel vectors

cannot be equal, unless theirmagnitudes equal 0.

16. Answers may vary. For example:and ,

and ,

and

17. As in question 15, non-parallel vectors.Their magnitudes must be 0 again tomake the equality true.

So, when , their sum will be 0.


1. a. false; Let , then:

b. true; and bothrepresent the lengths of thediagonal of a parallelogram, thefirst with sides and and thesecond with sides and sinceboth parallelograms have as aside and diagonals of equal length

c. true; Subtracting from both sidesshows that .b

!� c!a!@b! @ � @c! @ .

a!c!;a

! b!

a!

@a!� c! @@a!� b

! @� 0 6 0a! 0� 00 00a!� b! 0 � 0a!� 1�a

!2 0b!� �a

!� 0

m � �3�3m � 2,

m2 � m � 6 � 1m � 2 2 1m � 3 2�3m � 1,m2 � 2m � 3 � 1m � 1 2 1m � 3 2

q �2

3p �

13

3

q � 0p � 25

q � 1p � �6

n � 3m � 2,�7

� 1�3, 14, 7 231�1, 3, 4 2 � 510, �1, 1 2b14, 1, �2 2 � 1�14, �1, 16 21�1, 2, 3 2 �

� 181�1, 1 214, �11 2 � �712, 1 2� 1141�1, 1 21124, �5 2 � 11912, �1 212, �3 2 � �112, �1 2 � 41�1, 1 2b � x � 2ya � x � y

� 811, 5 2� 2 1�1, 321�10, �342 c � 3b � 24a � �2

�211, 0, 0 2 � 410, 1, 0 2

13, 4, 0 2 � 311, 1, 0 2 � 10, 1, 0 21�1, 2, 0 2 � �111, 1, 0 2 � 310, 1, 0 25 11, 1, 0 2 , 10, 1, 0 2 613, 4, 0 2 � 311, 0, 0 2 � 410, 1, 0 21�1, 2, 0 2 � �111, 0, 0 2 � 210, 1, 0 25 11, 0, 0 2 , 10, 1, 0 2 6:�7i!� 23j

!� 14k

!14i!� 43j

!� 40k

!R25 1�1, 1 2 , 1�3, 6 2 6 5 12, �4 2 , 1�1, 1 2 6,5 1�1, 2 2 , 1�1, 1 2 6,0!

1�21, 0, 0 2m11, 0, 0 2 .i

!m10, 1 2 � 10, �1 2 . m � �1,m10, 1 2 .

R2.

0!

V8 � 1�2, 11, 5 2V7 � 10, 9, 0 2V6 � 1�2, 7, 4 2V5 � 1�2, 6, 6 2V4 � 10, 5, �1 2V3 � 10, 4, 1 2V2 � 1�2, 2, 5 2V1 � 10, 0, 0 2

z

x

y

A

BO

C

OA

OB

OC

c � 0b � 7a �2

3

BC!� AD

!,

AB!� DC

!� 13, �4, 12 2DC!� 13, �4, 12 2AB!

BC!� AD

!.AB

!� DC

!

1�5, �2, �9 2

AB!10, �b, c 2�OB

!� � �b2 � c2

�OB!� � �a2 � c2

�OA!� � �a2 � b2

OC!� 0i!� bj!� ck!OB

!� ai!� 0j!� ck!OA

!� ai!� bj!� 0k!

10, b, c 2 1a, b, 0 2 , 1a, 0, c 2

OC!

OB!

OA!,

y!� �2i

!� 2j!� 6k!x

!� i!� 4j!� k!11.18

5.391.415.10

�9i!� 3j!� 3k!

9i!� 3j!� 3k!

3i!� 0j!� 0k!

i!� 2j!� 2k!

12, 30, �13 2a�13

2, 2,

3

2b

1�7, �16, 8 211, �3, 3 2OB!.OA

!AB!�AB!� � 14.07

AB!� 15, �2, �13 2�OP!� � 12.57

�OB!� � 3�OA

!� � 13

1�1, 6, 11 2�OB!� � 6.40OB

!� 13, 4, �4 24.58

�1i!� 2j!� 4k!


d. true; Draw the parallelogramformed by and and are the opposite sides of aparallelogram and must be equal.

e. true; the distributive law for scalarsf. false; Let and let

Then,and

but

so2. a.

b.3. a. ,

b.

4. a.

b.

5.

6. a. ,

b.

7. a. ,

,

b. 12.5c. 18.13d.

8. a.

b. 5

9. ,

,

10. a. where P(x, y, z)is the point.

b. (0, 0, 0) and

11. a. , ,

b. , ,

12. a. yesb. yes

13. a.

Since thetriangle is right-angled.

b.

14. a. and

b. and

c.

But

Therefore,15. a. C(3, 0, 5), P(3, 4, 5), E(0, 4, 5),

F(0, 4, 0)b. ,

c.d.

16. a. 7.74b. 2.83c. 2.83

17. a. 1236.9 kmb. S14.0°W

18. a. Any pair of nonzero, noncollinearvectors will span To show that(2, 3) and (3, 5) are noncollinear,show that there does not exist anynumber k such that .Solve the system of equations:

Solving both equations gives two

different values for k, and so (2, 3) and (3, 5) are noncollinearand thus span .

b. ,19. a. Find a and b such that

i.ii.iii.Use the method of elimination with i. and iii.

By substitution, .

lies in the plane determined by and because it can be written as alinear combination of and

b. If vector is in the span of and then can be written as a linearcombination of and Find m andn such that

Solve the system of equations:

Use the method of elimination:

By substitution,So, vector is in the span of and

20. a.

b.c.d. (4, 4, 0)

21. 722. a. ,

,

b. If A, B, and C are vertices of a righttriangle, then

So, triangle ABC is a right triangle.

23. a.b.c.d.e. b!� c!0

!�b!� a!� c!a

!� b!a

!� b!� c!

� 100�AB!�2 � 102

� 100� 20 � 80

�BC!�2 � �CA

!�2 � 12�5 22 � 1�8022�BC

!�2 � �CA

!�2 � �AB

!�2

�CA!� � �80 � 8.94

�BC!� � 2�5 � 4.47

�AB!� � 10

1�4, 0, �4 21�4, �4, �4 2

y

z

x

(0, 0, 0)

(0, 0, 4)

(0, 4, 0)

(4, 0, 0) (4, 4, 0)

(4, 0, 4)

(0, 4, 4)

(4, 4, 4)

c!.

b!

a! m � 11.

3 � n 33 � 11n

� �13 � �2m � 3n 46 � 2m � 8n

2123 2 � 21m � 4n 2 23 � m � 4n 36 � 3m � n

�13 � �2m � 3n

3m � n, m � 4n 2� 1�2m � 3n,� 13n, n, 4n 2� 1�2m, 3m, m 2� n13, 1, 4 21�13, 36, 23 2 � m1�2, 3, 1 2

c!.b

!a! c

!,b

!a!

c!.b

!c! b

!a!

a � 2

3 � b 33 � 11b

� 5 � �2a � 3b 28 � 2a � 8b

2114 2 � 21a � 4b 214 � a � 4b9 � 3a � b5 � �2a � 3b

� b, a � 4b 215, 9, 14 2 � 1�2a � 3b, 3a� 13b, b, 4b 215, 9, 14 2 � 1�2a, 3a, a 2� b13, 1, 4 215, 9, 14 2 � a1�2, 3, 1 2n � 621m � �770

R2

53,3

2

3k � 52k � 3

k12, 3 2 � 13, 5 2R2.

50.2°90°CF!� 1�3, 4, �5 2DB!� 13, 4, �5 2

0AD! 0 2 � 0DC

! 0 2 � 0DB! 0 20AC

! 0 2 � 0DB! 0 20AD

! 0 2 � 0DC! 0 2 � 0AC

! 0 2EA!

CE!,AB

!DC!,

ED!

EB!,BC

!DA!,

�6

3

�AB!�2 � �AC

!�2 � �BC

!�2

�AB!�2 � 9, �AC

!�2 � 3, �BC

!�2 � 6

c � �10b �7

3a � 8

c � 10b � 26.5a � �3

a1,1

3, 0 b

x � 3y � 6z � 031�3, 1 2 � 21�1, 2 2 � 1�11, 7 21

31�11, 7 2 � a�2

3b 1�1, 22� 1�3, 12

1

2 1�11, 7 2� a�3

2 b 1�3, 1 2� 1�1, 2 2

c

b

a

a – b a – ba – b + c

16, 2, �2 2�AB!� 2 � �CA

!�2 � �BC

!� 2

�CA!� � �45

�BC!� � �59

�AB!� � �14

u � 84.4°OA!� OB

!� 19, �4, �4 2OA

!� OB

!� 1�3, 8, �8 2

a�6

7,

2

7,

3

7b

a�2

3, �

1

3, �

2

3b

1�6, �3, �6 2a�

2

7,

3

7,

6

7b

0XY! 0 � 7

XY!� 1�2, 3, 6 2a

!� 3b

!� 3c!20a

!� 30b

!� 8c!0a!� b

! 0 � 0c!� d! 00c!� b

! 0 � 0c!� c! 0 � 02c

! 00a!� b! 0 � 0a!� 1�a

!2 0 � 00c! 0 � 0d! 00a! 0 � 0�a

! 0 � 0b! 0c!� d!� 0.

b!� �a

!

RS!

FW!

SW!.RF

!


24.

25. a.

b.

c.26. Case 1 If and are collinear, then

is also collinear with both and . But is perpendicular to and

, so is perpendicular to .Case 2 If and are not collinear,then by spanning sets, and span aplane in R3, and is in thatplane. If is perpendicular to and ,then it is perpendicular to the planeand all vectors in the plane. So, isperpendicular to .

Chapter 6 Test, p. 3481. Let P be the tail of and let Q be

the head of The vector sumsand can

be depicted as in the diagram below,using the triangle law of addition.

We see that

This is the associativeproperty for vector addition.

2. a. (8, 4, 8)b. 12

c.

3.

4. a. ,b. , ,

5. a. and span because anyvector (x, y) in can be written asa linear combination of and These two vectors are not multiplesof each other.

b. ,

6. a.

b. cannot be written as a linearcombination of and In otherwords, does not lie in the planedetermined by and

7. , from toward

8.

Also,

Thus,

Chapter 7

Review of Prerequisite Skills,p. 350

1. km h N E2. 15.93 units W N3.

4. a.b.c.d.

5. a. on the xy-planeb. on the xz-planec. on the yz-plane

6. a.

b.c.

d.

7. a.

b.

c.

Section 7.1, pp. 362–364

1. a. 10 N is a melon, 50 N is a chair,100 N is a computer

b. Answers will vary.2. a.

b.3. a line along the same direction

4. For three forces to be in equilibrium,they must form a triangle, which is aplanar figure.

5. a. The resultant is 13 N at an angle ofN E. The equilibrant is 13 Nat an angle of S W.

b. The resultant is 15 N at an angle ofS W. The equilibrant is 15 N atN E.

6. a. yes b. yes c. no d. yes7. Arms 90 cm apart will yield a resultant

with a smaller magnitude than at 30 cm apart. A resultant with a smallermagnitude means less force to counteryour weight, hence a harder chin-up.

8. The resultant would be 12.17 N atfrom the 6 N force toward the

8 N force. The equilibrant would be12.17 N at from the 6 N forceaway from the 8 N force.

9. 9.66 N from given force, 2.95 Nperpendicular to 9.66 N force

10. 49 N directed up the ramp11. a.

b.12. approximately south of east13. a. 7 N

b. The angle between and theresultant is The angle between

and the equilibrant is .14. a.

For these three equal forces to be inequilibrium, they must form anequilateral triangle. Since theresultant will lie along one of theselines, and since all angles of anequilateral triangle are theresultant will be at a angle withthe other two vectors.

b. Since the equilibrant is directed

60°60°,

1 N

1 N1 N

60°60°

60°

163.7°f!1

16.3°.f1

7.1 N 45°60°

13 N

7 N

8 N

15°

145.3°

34.7°

36.9°36.9°

22.6°22.6°

180°

10 N

20 N

30 N

12i!� j!� 2k!5i

!� j!� k!i

!� 3j!� k!4i

!� 6j!� 8k!�8i

!� 11j

!� 3k!6i

!� 2j!i

!� 7j!

l � 9.2212, 0, �9 2 ;l � 1.4111, 1, 0 2 ; l � 16.911�9, 3, 14 2 ;13, �2, 7 2 ; l � 7.87

yA(0, 1, 0)

z

x

B(–3, 2, 0)

C(–2, 0, 1)

D(0, 2, –3)

32.2°7.1°>v � 806

DE!�

1

2BA!

BA!� 2b

!� 2a

!BA!� CA

!� CB

!

DE!� b!� a!DE

!� CE

!� CD

!

y!

x!

u � 3.61; 73.9°�13

q!.p

!r! q

!.p

!r!711, 2,�3 211, 12, �29 2 � �213, 1, 4 2 �

q � 3p � �2

b!.a

!R2R2,b

!a! c � �11b � 5a � 1

� 3b!� 5a

!y!

x!� 2b

!� 3a

!�19

a�2

3, �

1

3, �

2

3b

(b + c)

(a + b)b

a

= (a + b) + c = a + (b + c)PQ

Q

P

c

1a!� b!2 � c

!.

PQ!� a!� 1b!� c

!2 �

3 1a!� b!2 � c

!43a!� 1b!� c!2 4c!.

a!

2b! � 4c

!

a!

c! b

! a

!

2b! � 4c

!

c! b

!

c! b

!

2b! � 4c

! a

! c

!

b! a

! c

!

b! 2b

! � 4c

!

c! b

!

�4 0a! 0 2 � 9 0b! 0 2� 0a! 0 2 � 0b! 0 2� 0a! 0 2 � 0b! 0 2 a

b


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